(a)

\begin{align} {dy \over dx} & = 2(2)(x) - 8(1) + 0 \\ & = 4x - 8 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4x - 8 \\ 8 & = 4x \\ {8 \over 4} & = x \\ 2 & = x \\ \\ {d^2y \over dx^2} & = {d \over dx}(4x - 8) \\ & = 4 \\ \\ \therefore \text{When } x = 2, & \text{ value of } y \text{ is a minimum} \end{align}

(b)

\begin{align} y & = x^2 + (2x - 1)^2 \\ \\ {dy \over dx} & = 2x + 2(2x - 1). (2) \\ & = 2x + 4(2x - 1) \\ & = 2x + 8x - 4 \\ & = 10x - 4 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 10x - 4 \\ 4 & = 10x \\ {4 \over 10} & = x \\ {2 \over 5} & = x \\ \\ {d^2 y \over dx^2} & = 10(1) \\ & = 10 \\ \\ \therefore \text{When } x = {2 \over 5}, & \text{ value of } y \text{ is a minimum} \end{align}

(c)

(d)

(e)

\begin{align} 2x + y & = 10 \\ y & = 10 - 2x \\ \\ A & = xy \\ \\ \text{Since } & y = 10 - 2x, \\ A & = x(10 - 2x) \\ & = 10x - 2x^2 \\ \\ {dA \over dx} & = 10(1) - 2(2)(x) \\ & = 10 - 4x \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 10 - 4x \\ 4x & = 10 \\ x & = {10 \over 4} \\ & = 2.5 \\ \\ {dA \over dx} & = 10 - 4x \\ \\ {d^2 A \over dx^2} & = - 4(1) \\ & = -4 \implies A \text{ is a maximum} \\ \\ \text{Substitute } & x = 2.5 \text{ into } A = 10x - 2x^2, \\ A & = 10(2.5) - 2(2.5)^2 \\ & = 12.5 \\ \\ \therefore \text{Maximum } & \text{value of } A = 12.5 \end{align}

(i) Note: The fencing is 36 m long and it doesn’t include the wall!

\begin{align} \text{Length of fencing } & = PS + SR + RQ \\ 36 & = x + y + x \\ 36 & = 2x + y \\ 36 - 2x & = y \\ \\ y & = 36 - 2x \phantom{000} \text{(Shown)} \end{align}

(ii)

\begin{align} A & = xy \\ \\ \text{Since } & y = 36 - 2x, \\ A & = x(36 - 2x) \\ & = 36x - 2x^2 \text{ (Shown)} \end{align}

(iii)

\begin{align} {dA \over dx} & = 36(1) - 2(2)(x) \\ & = 36 - 4x \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 36 - 4x \\ 4x & = 36 \\ x & = {36 \over 4} \\ & = 9 \\ \\ {dA \over dx} & = 36 - 4x \\ \\ {d^2 A \over dx^2} & = -4(1) \\ & = -4 \\ \\ \therefore \text{When } & x = 9, \phantom{.} A \text{ is a maximum} \end{align}

(i)

\begin{align} \text{Volume} & = \text{Length } \times \text{Breadth } \times \text{Height} \\ 400 & = x \times 8 \times \text{Height} \\ 400 & = 8x \times \text{Height} \\ {400 \over 8x} & = \text{Height} \\ {50 \over x} & = \text{Height} \\ \\ \text{Height} & = {50 \over x} \text{ cm} \end{align}

(ii) Note: It is an open rectangular box, thus do not calculate the area of the 'lid'

\begin{align} A & = \left(8 \times x\right) + 2\left(8 \times {50 \over x}\right) + 2\left(x \times {50 \over x}\right) \\ & = 8x + 2\left(400 \over x\right) + 2(50) \\ & = 8x + {800 \over x} + 100 \phantom{00} \text{ (Shown)} \end{align}

(iii)

\begin{align} A & = 8x + {800 \over x} + 100 \\ & = 8x + 800x^{-1} + 100 \\ \\ {dA \over dx} & = 8(1) + 800(-1)x^{-2} + 0 \\ & = 8 - 800x^{-2} \\ & = 8 - 800 \left(1 \over x^2\right) \\ & = 8 - {800 \over x^2} \\ \\ {d^2 A \over dx^2} & = - 800(-2)x^{-3} \\ & = 1600x^{-3} \\ & = {1600 \over x^3} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 8 - {800 \over x^2} \\ {800 \over x^2} & = 8 \\ 800 & = 8x^2 \\ {800 \over 8} & = x^2 \\ 100 & = x^2 \\ \pm\sqrt{100} & = x \\ \pm 10 & = x \\ \\ x & = 10 \text{ or } - 10 \text{ (N.A.)} \\ \\ \text{Substitute } & x = 10 \text{ into } {d^2 A \over dx^2}, \\ {d^2 A \over dx^2} & = {1600 \over (10)^3} \\ & = 1.6 \\ \\ \therefore \text{When } & x = 10, \phantom{.} A \text{ is a minimum} \end{align}

(i)

\begin{align} \text{Perimeter of trapezium } & = y + 5x + (6x + y) + 5x \\ 104 & = 2y + 16x \\ 104 - 16x & = 2y \\ 52 - 8x & = y \\ \\ y & = 52 - 8x \end{align}

Since the area of trapezium is found by ${1 \over 2} (a + b) (h)$, we need to find the height of the trapezium:

Since the non-parallel sides are equal in length, this is an isosceles trapezium.

\begin{align} \text{By Py} & \text{thagora's theorem,} \\ (5x)^2 & = h^2 + (3x)^2 \\ 25x^2 & = h^2 + 9x^2 \\ 16x^2 & = h^2 \\ \pm \sqrt{16x^2} & = h \\ \pm 4x & = h \\ \\ h & = 4x \text{ or } -4x \text{ (N.A.)} \\ \\ A & = {1 \over 2} \times \text{Sum of parallel sides } \times \text{Height} \\ & = {1 \over 2} \times [y + (6x + y)] \times 4x \\ & = {1 \over 2}(6x + 2y)(4x) \\ & = (3x + y)(4x) \\ & = 12x^2 + 4xy \\ \\ \text{Since } & y = 52 - 8x, \\ A & = 12x^2 + 4x(52 - 8x) \\ & = 12x^2 + 208x - 32x^2 \\ & = 208x - 20x^2 \phantom{000} \text{ (Shown)} \end{align}

(ii)

\begin{align} {dA \over dx} & = 208(1) - 20(2)(x) \\ & = 208 - 40x \\ \\ {d^2 A \over dx^2} & = -40(1) \\ & = -40 \implies A \text{ is a maximum} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 208 - 40x \\ 40x & = 208 \\ x & = {208 \over 40} \\ & = 5.2 \\ \\ \text{Substitute } & x = 5.2 \text{ into } y = 52 - 8x, \\ y & = 52 - 8(5.2) \\ & = 10.4 \\ \\ \therefore \text{When } x = 5.2 & \text{ and } y = 10.4, \phantom{.} A \text{ is a maximum} \end{align}

(i)

\begin{align} x^4 y & = 32 \\ y & = {32 \over x^4} \\ \\ z & = x^2 + y \\ \\ \text{Since } & y = {32 \over x^4}, \\ z & = x^2 + {32 \over x^4} \phantom{000} \text{ (Shown)} \end{align}

(ii)

\begin{align} z & = x^2 + {32 \over x^4} \\ & = x^2 + 32x^{-4} \\ \\ {dz \over dx} & = 2x + 32(-4)x^{-5} \\ & = 2x - 128 x^{-5} \\ & = 2x - {128 \over x^5} \\ \\ {d^2 z \over dx^2} & = 2(1) - 128(-5)x^{-6} \\ & = 2 + 640x^{-6} \\ & = 2 + {640 \over x^6} \\ \\ \text{Let } & {dz \over dx} = 0, \\ 0 & = 2x - {128 \over x^5} \\ {128 \over x^5} & = 2x \\ 128 & = x^5 (2x) \\ 128 & = 2x^6 \\ {128 \over 2} & = x^6 \\ 64 & = x^6 \\ \pm \sqrt[6]{64} & = x \\ \pm 2 & = x \\ \\ x & = 2 \text{ or } -2 \text{ (Reject, since } x > 0) \\ \\ \text{Substitute } & x = 2 \text{ into } {d^2 z \over dx^2}, \\ {d^2 z \over dx^2} & = 2 + {640 \over (2)^6} \\ & = 12 \\ \\ \therefore \text{When } & x = 2, \phantom{.} z \text{ is a minimum} \end{align}

Recall that the tangent and normal to the curve at the same point are perpendicular, i.e. $m_1 \times m_2 = -1$

(i) Note: For a cylinder, the volume can be found by $\pi r^2 h$ and the curved surface area can be found by $2 \pi r h$

\begin{align} \text{Volume } & = \pi r^2 h \\ 400 & = \pi r^2 \times h \\ {400 \over \pi r^2} & = h \\ \\ h & = {400 \over \pi r^2} \\ \\ \\ A & = 2\pi rh + 2 \times \pi r^2 \\ & = 2\pi rh + 2\pi r^2 \\ \\ \text{Since } & h = {400 \over \pi r^2}, \\ A & = 2\pi r\left( 400 \over \pi r^2 \right) + 2\pi r^2 \\ & = {800\pi r \over \pi r^2} + 2\pi r^2 \\ & = {800 \over r} + 2\pi r^2 \phantom{000} \text{ (Shown)} \end{align}

(ii)

\begin{align} A & = {800 \over r} + 2\pi r^2 \\ & = 800r^{-1} + 2\pi r^2 \\ \\ {dA \over dr} & = 800(-1)r^{-2} + 2\pi (2) r \\ & = - 800 r^{-2} + 4\pi r \\ & = -{800 \over r^2} + 4\pi r \\ \\ {d^2 A \over dr^2} & = -800(-2)r^{-3} + 4\pi(1) \\ & = 1600r^{-3} + 4 \pi \\ & = {1600 \over r^3} + 4\pi \\ \\ \text{Let } & {dA \over dr} = 0, \\ 0 & = -{800 \over r^2} + 4\pi r \\ {800 \over r^2} & = 4\pi r \\ 800 & = 4\pi r(r^2) \\ 800 & = 4\pi r^3 \\ {800 \over 4\pi} & = r^3 \\ {200 \over \pi} & = r^3 \\ r^3 & = {200 \over \pi} \\ r & = \sqrt[3]{200 \over \pi} \\ & = 3.993 \\ \\ \text{Substitute } & r = 3.993 \text{ into } {d^2 A \over dr^2}, \\ {d^2 A \over dr^2} & = {1600 \over (3.993)^3} + 4\pi \\ & = 37.699 \\ \\ \therefore \text{When } r \approx 3.99 \text{ cm}, & \text{ the surface area of the cylinder is a minimum} \end{align}

(i) Note: It is an open rectangular box, thus do not calculate the area of the 'lid'

\begin{align} \text{Volume } & = \text{Base area } \times \text{Height} \\ 36 & = (x)(2x) \times h \\ 36 & = 2x^2 \times h \\ {36 \over 2x^2} & = h \\ {18 \over x^2} & = h \\ \\ h & = {18 \over x^2} \\ \\ \\ A & = (2x)(x) + 2(h)(x) + 2(h)(2x) \\ & = 2x^2 + 2hx + 4hx \\ & = 2x^2 + 6hx \\ \\ \text{Since } & h = {18 \over x^2}, \\ A & = 2x^2 + 6\left(18 \over x^2 \right)x \\ & = 2x^2 + \left(108x \over x^2\right) \\ & = 2x^2 + {108 \over x} \phantom{000} \text{ (Shown)} \end{align}

(ii)

\begin{align} A & = 2x^2 + {108 \over x} \\ & = 2x^2 + 108x^{-1} \\ \\ {dA \over dx} & = 2(2)(x) + 108(-1)(x^{-2}) \\ & = 4x - 108 x^{-2} \\ & = 4x - {108 \over x^2} \\ \\ {d^2 A \over dx^2} & = 4(1) - 108(-2)x^{-3} \\ & = 4 + 216 x^{-3} \\ & = 4 + {216 \over x^3} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 4x - {108 \over x^2} \\ {108 \over x^2} & = 4x \\ 108 & = 4x(x^2) \\ 108 & = 4x^3 \\ {108 \over 4} & = x^3 \\ 27 & = x^3 \\ \sqrt[3]{27} & = x \\ 3 & = x \\ \\ \text{Substitute } & x = 3 \text{ into } h = {18 \over x^2}, \\ h & = {18 \over (3)^2} \\ & = 2 \\ \\ \text{Substitute } & x = 3 \text{ into } {d^2 A \over dx^2}, \\ {d^2A \over dx^2} & = 4 + {216 \over (3)^3} \\ & = 12 \\ \\ \therefore \text{When } x = 3 \text{ and } h = 2, & \text{ the total surface area is a minimum} \end{align}

Show part

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ ST^2 & = TM^2 + SM^2 \\ & = (x)^2 + \left(3x \over 4\right)^2 \\ & = x^2 + {9x^2 \over 16} \\ & = {16x^2 \over 16} + {9x^2 \over 16} \\ & = {25x^2 \over 16} \\ \\ ST & = \pm \sqrt{25x^2 \over 16} \\ & = \pm {5x \over 4}\\ & = {5x \over 4} \text{ or } -{5x \over 4} \text{ (N.A.)} \\ \\ \text{Perimeter } & = PQ + QR + RS + ST + TP \\ 30 & = 2x + y + {5x \over 4} + {5x \over 4} + y \\ 30 & = 2x + {5x \over 4} + {5x \over 4} + y + y \\ 30 & = {9x \over 2} + 2y \\ 30 - {9x \over 2} & = 2y \\ 15 - {9x \over 4} & = y \\ \\ y & = 15 - {9x \over 4} \\ \\ A & = \text{Area of rectangle } PQRT + \text{ Area of triangle }RST \\ & = (2x)(y) + {1 \over 2}(2x)\left(3x \over 4 \right) \\ & = 2xy + (x)\left(3x \over 4\right) \\ & = 2xy + {3x^2 \over 4} \\ \\ \text{Since } & y = 15 - {9x \over 4}, \\ A & = 2x\left( 15 - {9x \over 4} \right) + {3x^2 \over 4} \\ & = 30x - {9x^2 \over 2} + {3x^2 \over 4} \\ & = 30x - {15x^2 \over 4} \phantom{000} \text{ (Shown)} \end{align}

(i)

\begin{align} A & = 30x - {15x^2 \over 4} \\ \\ {dA \over dx} & = 30(1) - {15(2)(x) \over 4} \\ & = 30 - {30x \over 4} \\ \\ {d^2 A \over dx^2} & = - {30(1) \over 4} \\ & = -7.5 \implies A \text{ is a maximum} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 30 - {30x \over 4} \\ {30x \over 4} & = 30 \\ 30x & = 4(30) \\ 30x & = 120 \\ x & = {120 \over 30} \\ & = 4 \\ \\ \text{Substitute } & x = 4 \text{ into } A = 30x - {15x^2 \over 4} \\ A & = 30(4) - {15(4)^2 \over 4} \\ & = 60 \\ \\ \therefore \text{Maximum} & \text{ value of } A = 60 \text{ cm}^2 \end{align}

(ii)

\begin{align} SN & = SM + MN \\ & = {3x \over 4} + y \\ \\ \text{Since } & y = 15 - {9x \over 4}, \\ SN & = {3x \over 4} + \left(15 - {9x \over 4} \right) \\ & = 15 + {3x \over 4} - {9x \over 4} \\ & = 15 - {6x \over 4} \\ & = 15 - {3x \over 2} \\ \\ \text{When } & x = 4, \\ SN & = 15 - {3(4) \over 2} \\ & = 9 \text{ cm} \end{align}

(i) Trigonometry formulas used (Cosine rule and area of triangle) can be found in the provided formula sheet

\begin{align} \text{By Co} & \text{sine rule,} \\ a^2 & = b^2 + c^2 - 2bc\cos A \\ BC^2 & = AC^2 + AB^2 - 2(AC)(AB)\cos \angle BAC \\ (6x)^2 & = (5x)^2 + (5x)^2 - 2(5x)(5x) \cos \angle BAC \\ 36x^2 & = 25x^2 + 25x^2 - 50x^2 \cos \angle BAC \\ 36x^2 - 25x^2 - 25x^2 & = -50x^2 \cos \angle BAC \\ -14x^2 & = -50x^2 \cos \angle BAC \\ {-14x^2 \over -50x^2} & = \cos \angle BAC \\ {7 \over 25} & = \cos \angle BAC \\ \\ \cos \angle BAC & = {7 \over 25} \\ \angle BAC & = \cos^{-1} \left(7 \over 25\right) \\ \\ \text{Area of } \triangle BAC & = {1 \over 2}ab \sin C \\ & = {1 \over 2}(AB)(AC)\sin \angle BAC \\ & = {1 \over 2}(5x)(5x)\sin \left[ \cos^{-1} \left(7 \over 25\right) \right] \\ & = (12.5x^2)(0.96) \\ & = 12x^2 \text{ cm}^2 \\ \\ \text{Volume } & = \text{Cross-sectional area } \times \text{Height} \\ 4500 & = \text{Area of }\triangle BAC \times h \\ 4500 & = 12x^2 \times h \\ {4500 \over 12x^2} & = h \\ {375 \over x^2} & = h \\ \\ h & = {375 \over x^2} \end{align}

(ii)

\begin{align} A & = 2\times \text{Area of }\triangle BAC + \text{Area of } ABED + \text{Area of }BCFE + \text{Area of }ACFD \\ & = 2\times 12x^2 + (5x)(h) + (6x)(h) + (5x)(h) \\ & = 24x^2 + 16xh \\ \\ \text{Since } & h = {375 \over x^2}, \\ A & = 24x^2 + 16x\left(375 \over x^2\right) \\ & = 24x^2 + {6000x \over x^2} \\ & = 24x^2 + {6000 \over x} \phantom{000} \text{ (Shown)} \end{align}

(iii)

\begin{align} A & = 24x^2 + {6000 \over x} \\ & = 24x^2 + 6000x^{-1} \\ \\ {dA \over dx} & = 24(2)x + 6000 (-1)x^{-2} \\ & = 48x - 6000 x^{-2} \\ & = 48x - {6000 \over x^2} \end{align}

(iv)

\begin{align} \text{Let } & {dA \over dx} = 0, \\ 0 & = 48x - {6000 \over x^2} \\ {6000 \over x^2} & = 48x \\ 6000 & = 48x(x^2) \\ 6000 & = 48x^3 \\ {6000 \over 48} & = x^3 \\ 125 & = x^3 \\ \sqrt[3]{125} & = x \\ 5 & = x \\ \\ \text{Substitute } & x = 5 \text{ into } A = 24x^2 + {6000 \over x}, \\ A & = 24(5)^2 + {6000 \over (5)} \\ & = 1800 \end{align}

(a)

\begin{align} \text{Let } x \text{ denote } & \text{the breadth of the rectangle} \\ \text{Let } y \text{ denote } & \text{the length of the rectangle} \\ \text{Let } P \text{ denote } & \text{the perimeter of the rectangle} \\ \text{Let } A \text{ denote } & \text{the area of the rectangle} \\ \\ P & = 2\times \text{Length} + 2 \times \text{Breadth} \\ 80 & = 2 \times y + 2 \times x \\ 80 & = 2y + 2x \\ 40 & = y + x \\ 40 - x & = y \\ \\ y & = 40 - x \\ \\ A & = \text{Length} \times \text{Breadth} \\ & = y \times x \\ & = yx \\ \\ \text{Since } & y = 40 - x, \\ A & = (40 - x)x \\ & = 40x - x^2 \\ \\ {dA \over dx} & = 40(1) - (2)(x) \\ & = 40 - 2x \\ \\ {d^2 A \over dx^2} & = -2(1) \\ & = -2 \implies A \text{ is a maximum} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 40 - 2x \\ 2x & = 40 \\ x & = {40 \over 2} \\ & = 20 \\ \\ \text{Substitute } & x = 20 \text{ into } A = 40x - x^2, \\ A & = 40(20) - (20)^2 \\ & = 400 \\ \\ \therefore \text{Maximum} & \text{ area} = 400 \text{ m}^2 \end{align}

(b)

\begin{align} \text{Let } x \text{ denote } & \text{the breadth of the rectangle} \\ \text{Let } y \text{ denote } & \text{the length of the rectangle} \\ \text{Let } P \text{ denote } & \text{the perimeter of the rectangle} \\ \text{Let } A \text{ denote } & \text{the area of the rectangle} \\ \\ A & = \text{Length} \times \text{Breadth} \\ 16 & = y \times x \\ 16 & = yx \\ {16 \over x} & = y \\ \\ y & = {16 \over x} \\ \\ P & = 2 \times \text{Length} + 2 \times \text{Breadth} \\ & = 2 \times y + 2 \times x \\ & = 2y + 2x \\ \\ \text{Since } & y = {16 \over x}, \\ P & = 2\left( 16 \over x \right) + 2x \\ & = {32 \over x} + 2x \\ & = 32x^{-1} + 2x \\ \\ {dP \over dx} & = 32(-1)x^{-2} + 2(1) \\ & = -32x^{-2} + 2 \\ & = -{32 \over x^2} + 2 \\ \\ {d^2 P \over dx^2} & = -32(-2)x^{-3} \\ & = 64x^{-3} \\ & = {64 \over x^3} \\ \\ \text{Let } & {dP \over dx} = 0, \\ 0 & = -{32 \over x^2} + 2 \\ {32 \over x^2} & = 2 \\ 32 & = 2x^2 \\ {32 \over 2} & = x^2 \\ 16 & = x^2 \\ \pm \sqrt{16} & = x \\ \pm 4 & = x \\ \\ x & = 4 \text{ or } - 4 \text{ (N.A.)} \\ \\ \text{Substitute } & x = 4 \text{ into } P = {32 \over x} + 2x, \\ P & = {32 \over (4)} + 2(4) \\ & = 16 \\ \\ \text{Substitute } & x = 4 \text{ into } {d^2 P \over dx^2}, \\ {d^2 P \over dx^2} & = {64 \over (4)^3} \\ & = 1 \\ \\ \therefore \text{Minimum} & \text{ perimeter} = 16 \text{ cm} \end{align}

\begin{align} \text{Let } W \text{ denote } & \text{the work done by the cell} \\ \\ W & = {E^2 R \over (r + R)^2} \end{align} \begin{align} u & = E^2 R &&& v & = (r + R)^2 \\ {du \over dR} & = E^2 &&& {dv \over dR} & = 2(r + R) . (1) \phantom{00000} [\text{Chain rule}] \\ & &&& & = 2(r + R) \end{align} \begin{align} {dW \over dR} & = { (r + R)^2 (E^2) - (E^2 R)[2(r + R)] \over [ (r + R)^2 ] ^2} \phantom{00000} [\text{Quotient rule}] \\ & = { E^2 (r + R)^2 - 2 E^2 R (r + R) \over (r + R)^4} \\ & = { E^2 (r + R) [ (r + R) - 2R ] \over (r + R)^4} \\ & = { E^2 (r + R) (r + R - 2R) \over (r + R)^4 } \\ & = { E^2 (r + R) (r - R) \over (r + R)^4 } \\ & = { E^2 (r - R) \over (r + R)^3} \\ \\ \text{Let } & {dW \over dR} = 0, \\ 0 & = {E^2 (r - R) \over (r + R)^3} \\ 0 & = E^2 (r - R) \\ {0 \over E^2} & = r - R \\ 0 & = r - R \\ R & = r \\ \\ {dW \over dR} & = { E^2 (r - R) \over (r + R)^3} \end{align}
\begin{align} u & = E^2 (r - R) &&& v & = (r + R)^3 \\ {du \over dR} & = E^2 ( - 1) &&& {dv \over dR} & = 3(r + R)^2 . (1) \\ & = - E^2 &&& & = 3(r + R)^2 \end{align} \begin{align} {d^2 W \over dR^2} & = { (r + R)^3 (-E^2) - E^2 (r - R)[3(r + R)^2] \over [(r + R)^3]^2 } \\ & = { - E^2 (r + R)^3 - 3 E^2 (r - R) (r + R)^2 \over (r + R)^6 } \\ & = { - E^2 (r + R)^2 [ (r + R) + 3 (r - R)] \over (r + R)^6 } \\ & = { - E^2 (r + R)^2 ( r + R + 3r - 3R ) \over (r + R)^6 } \\ & = { - E^2 (r + R)^2 (4r - 2R) \over (r + R)^6 } \\ & = { - E^2 (4r - 2R) \over (r + R)^4 } \\ \\ \text{When } & R = r, \\ {d^2 W \over dR^2} & = { - E^2 [4r - 2(r)] \over (r + r)^4 } \\ & = { - E^2 (4r - 2r) \over (2r)^4} \\ & = { - E^2 (2r) \over (2r)^4} \\ & = { - E^2 \over (2r)^3 } \\ \\ \text{Since } & E > 0 \text{ and } r > 0, {d^2 W \over dR^2} < 0 \\ \\ \therefore \text{Work } & \text{done is greatest when } R = r \end{align}

Note: The breadth of the rectangle is equals to the diameter of the semicircle

\begin{align} \text{Let } r \text{ denote } & \text{the radius of the semicircle} \\ \text{Let } l \text{ denote } & \text{the length of the rectangle} \\ \\ \text{Breadth of rectangle} & = 2r \\ \\ \text{Circumference of semicircle} & = {1 \over 2} (2\pi r) \\ & = \pi r \\ \\ \text{Perimeter} & = \pi r + l + l + 2r \\ 6 & = \pi r + 2l + 2r \\ 6 - \pi r - 2r & = 2l \\ \\ l & = {6 - \pi r - 2r \over 2} \\ & = 3 - {1 \over 2} \pi r - r \\ \\ \text{Let } A \text{ denote } & \text{the area of the window} \\ \\ A & = (l)(2r) + {1 \over 2} \pi r^2 \\ & = 2 l r + {1 \over 2} \pi r^2 \\ \\ \text{Since } & l = 3 - {1 \over 2} \pi r - r, \\ A & = 2r \left(3 - {1 \over 2} \pi r - r \right) + {1 \over 2} \pi r^2 \\ & = 6r - \pi r^2 - 2r^2 + {1 \over 2} \pi r^2 \\ & = 6r - {1 \over 2} \pi r^2 - 2r^2 \\ \\ {dA \over dr} & = 6(1) - {1 \over 2} \pi (2) r - 2(2)r \\ & = 6 - \pi r - 4r \\ \\ {d^2 A \over dr^2} & = - \pi (1) - 4(1) \\ & = - \pi - 4 \implies A \text{ is a maximum} \\ \\ \text{Let } & {dA \over dr} = 0, \\ 0 & = 6 - \pi r - 4r \\ 4 r + \pi r & = 6 \\ r (4 + \pi) & = 6 \\ r & = {6 \over 4 + \pi} \\ \\ \text{Substitute } & r = {6 \over 4 + \pi} \text{ into } A = 6r - {1 \over 2} \pi r^2 - 2r^2, \\ A & = 6 \left(6 \over 4 + \pi \right) - {1 \over 2} \pi \left(6 \over 4 + \pi \right)^2 - 2 \left(6 \over 4 + \pi \right)^2 \\ & = {36 \over 4 + \pi } - {\pi \over 2} \left[36 \over (4 + \pi)^2 \right] - 2 \left[36 \over (4 + \pi)^2 \right] \\ & = {36 \over 4 + \pi } - { 36 \pi \over 2(4 + \pi)^2 } - {72 \over (4 + \pi)^2 } \\ & = {36(4 + \pi) \over (4 + \pi)^2} - { 18 \pi \over (4 + \pi)^2 } - {72 \over (4 + \pi)^2} \\ & = {36(4 + \pi) - 18\pi - 72 \over (4 + \pi)^2} \\ & = {144 + 36 \pi - 18\pi - 72 \over (4 + \pi)^2} \\ & = {72 + 18\pi \over (4 + \pi)^2} \\ & = {18 (4 + \pi) \over (4 + \pi)^2} \\ & = {18 \over 4 + \pi} \text{ m}^2 \phantom{000} \text{ (Shown)} \end{align}

Note: The radius of the circle is fixed at $a$ cm - only the length and the breadth of the rectangle changes

\begin{align} \text{Let } l \text{ denote } & \text{the length of the rectangle} \\ \text{Let } b \text{ denote } & \text{the breadth of the rectangle} \\ \\ \text{By Pyth} & \text{agoras theorem,} \\ (2a)^2 & = l^2 + b^2 \\ 4a^2 & = l^2 + b^2 \\ \\ b^2 & = 4a^2 - l^2 \\ b & = \sqrt{4a^2 - l^2} \\ \\ \text{Let } A \text{ denote } & \text{the area of the rectangle} \\ \\ A & = (l)(b) \\ \\ \text{Since } & b = \sqrt{4a^2 - l^2}, \\ A & = l \sqrt{4a^2 - l^2} \\ & = l (4a^2 - l^2)^{1 \over 2} \end{align} \begin{align} u & = l &&& v & = (4a^2 - l^2)^{1 \over 2} \\ {du \over dl} & = 1 &&& {dv \over dl} & = {1 \over 2}(4a^2 - l^2)^{-{1 \over 2}} . (-2l)\\ & &&& & = -l (4a^2 - l^2)^{-{1 \over 2}} \\ & &&& & = {-l \over \sqrt{4a^2 - l^2}} \end{align} \begin{align} {dA \over dl} & = (l) \left( {-l \over \sqrt{4a^2 - l^2}} \right) + \left(\sqrt{4a^2 - l^2}\right) (1) \phantom{00000} [\text{Product rule}] \\ & = {-l^2 \over \sqrt{4a^2 - l^2}} + \sqrt{4a^2 - l^2} \\ & = {-l^2 \over \sqrt{4a^2 - l^2}} + {4a^2 - l^2 \over \sqrt{4a^2 - l^2}} \\ & = {4a^2 - 2l^2 \over \sqrt{4a^2 - l^2}} \end{align}
\begin{align} u & = 4a^2 - 2l^2 &&& v & = \sqrt{4a^2 - l^2} \\ & &&& & = (4a^2 - l^2)^{1 \over 2} \\ {du \over dl} & = -4l &&& {dv \over dl} & = {1 \over 2}(4a^2 - l^2)^{-{1 \over 2}} . (-2l) \\ & &&& & = -l (4a^2 - l^2)^{-{1 \over 2}} \\ & &&& & = {-l \over \sqrt{4a^2 - l^2}} \end{align} \begin{align} {d^2 A \over dl^2} & = { \left( \sqrt{4a^2 - l^2} \right) (-4l) - (4a^2 - 2l^2) \left({-l \over \sqrt{4a^2 - l^2}} \right) \over \left(\sqrt{4a^2 - l^2} \right)^2 } \\ & = { -4l \sqrt{4a^2 - l^2} + {4a^2 - 2l^2 \over \sqrt{4a^2 - l^2}} \over 4a^2 - l^2} \\ & = { -4l (4a^2 - l^2) + 4a^2 - 2l^2 \over (4a^2 - l^2) \sqrt{4a^2 - l^2} } \\ & = { -16 a^2 l + 4l^3 + 4a^2 - 2l^2 \over (4a^2 - l^2)^{3 \over 2} } \\ \\ \text{Let } & {dA \over dl} = 0, \\ 0 & = {4a^2 - 2l^2 \over \sqrt{4a^2 - l^2} } \\ 0 & = 4a^2 - 2l^2 \\ 2l^2 & = 4a^2 \\ l^2 & = 2a^2 \\ l & = \sqrt{2a^2} \\ & = \sqrt{2} a \\ \\ \text{Substitute } & l = \sqrt{2} a \text{ into } {d^2 A \over dl^2}, \\ {d^2 A \over dl^2} & = { - 16 a^2 (\sqrt{2} a) + 4(\sqrt{2} a)^3 + 4a^2 - 2(\sqrt{2} a)^2 \over [4a^2 - (\sqrt{2} a)^2 ]^{3 \over 2} } \\ & = { -16 \sqrt{2} a^3 + 4(2 \sqrt{2} a^3) + 4a^2 - 2(2a^2) \over (4a^2 - 2a^2)^{3 \over 2} } \\ & = { -16 \sqrt{2} a^3 + 8 \sqrt{2} a^3 + 4a^2 - 4a^2 \over (2a^2)^{3 \over 2} } \\ & = { -8 \sqrt{2} a^3 \over (2a^2)^{3 \over 2}} \\ \\ \text{Since } a > 0, & \phantom{.} {d^2 A \over dl^2} < 0 \text{ and } A \text{ is a maximum} \\ \\ \text{Substitute } & l = \sqrt{2} a \text{ into } b = \sqrt{4a^2 - l^2}, \\ b & = \sqrt{ 4a^2 - (\sqrt{2} a)^2 } \\ & = \sqrt{ 4a^2 - 2a^2 } \\ & = \sqrt{ 2a^2} \\ & = \sqrt{2} a \\ \\ \text{Since } l = b & = \sqrt{2} a \text{ cm, the rectangle is a square of side } \sqrt{2} a \text{ cm} \end{align}

\begin{align} \text{Since the point } & (x, y) \text{ lies on the parabola, point is } (x, 12 - x^2) \\ \\ \text{Let } A \text{ denote } & \text{the area of the rectangle} \\ \\ A & = 2(x)(y) \\ & = 2(x)(12 - x^2) \\ & = 2x(12 - x^2) \\ & = 24x - 2x^3 \\ \\ {dA \over dx} & = 24(1) - 2(3)x^2 \\ & = 24 - 6x^2 \\ \\ {d^2 A \over dx^2} & = - 6(2)x \\ & = -12x \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 24 - 6x^2 \\ 6x^2 & = 24 \\ x^2 & = {24 \over 6} \\ & = 4 \\ x & = \pm \sqrt{4} \\ & = \pm 2 \\ \\ x & = 2 \phantom{.} \text{ or } \phantom{.} - 2 \text{ (N.A.)} \\ \\ \text{Substitute } & x = 2 \text{ into } {d^2 A \over dx^2}, \\ {d^2 A \over dx^2} & = -12(2) \\ & = -24 \\ \\ \therefore \text{When } x = 2, & \phantom{.} A \text{ is a maximum} \\ \\ \text{Substitute } & x = 2 \text{ into } A = 24x - 2x^3, \\ A & = 24(2) - 2(2)^3 \\ & = 32 \text{ square units} \end{align}

\begin{align} & 1) \text{ Besides cost of the material, we need to consider the cost of sealing the lid to the can} \\ \\ & 2) \text{ The lid & base may be made smaller to reduce wastage} \end{align}