$$ \text{Let } y = (x - 2)\sqrt{5 - x} $$
\begin{align} u & = x - 2 &&& v & = \sqrt{5 - x} \\ & &&& & = (5 - x)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(5 - x)^{-{1 \over 2}}. (-1) \phantom{00000} [\text{Chain rule}] \\ & &&& & = -{1 \over 2}(5 - x)^{-{1 \over 2}} \\ & &&& & = -{1 \over 2\sqrt{5 - x}} \end{align} \begin{align} {dy \over dx} & = (x - 2)\left(-{1 \over 2\sqrt{5 - x}}\right) + (\sqrt{5 - x})(1) \phantom{00000} [\text{Product rule}] \\ & = {-(x - 2) \over 2\sqrt{5 - x}} + \sqrt{5 - x} \\ & = {2 - x \over 2\sqrt{5 - x}} + \sqrt{5 - x} \times {2\sqrt{5 - x} \over 2\sqrt{5 - x}} \\ & = {2 - x \over 2\sqrt{5 - x}} + {2(5 - x) \over 2\sqrt{5 - x}} \\ & = {2 - x + 2(5 - x) \over 2\sqrt{5 - x}} \\ & = {2 - x + 10 - 2x \over 2\sqrt{5 - x}} \\ & = {12 - 3x \over 2\sqrt{5 - x}} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {12 - 3x \over 2\sqrt{5 - x}} \\ 0 & = 12 - 3x \\ 3x & = 12 \\ x & = {12 \over 3} \\ & = 4 \\ \\ \text{Substitute } & x = 4 \text{ into eqn of } y, \\ y & = ((4) - 2)\sqrt{5 - (4)} \\ & = 2 \\ \\ \therefore \text{Stationary} & \text{ value} = 2 \end{align}

(i)

\begin{align} y & = 4x^2 + {27 \over x} \\ & = 4x^2 + 27x^{-1} \\ \\ {dy \over dx} & = 4(2)(x) + 27(-1)(x^{-2}) \\ & = 8x - 27x^{-2} \\ & = 8x - {27 \over x^2} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 8x - {27 \over x^2} \\ {27 \over x^2} & = 8x \\ 27 & = 8x^3 \\ {27 \over 8} & = x^3 \\ \sqrt[3]{27 \over 8} & = x \\ {3 \over 2} & = x \\ \\ \text{Substitute } & x = {3 \over 2} \text{ into eqn of curve,} \\ y & = 4\left(3 \over 2\right)^2 + {27 \over ({3 \over 2})} \\ & = 27 \\ \\ \therefore \text{Turning} & \text{ point is } \left({3 \over 2}, 27 \right) \end{align}

(ii)

\begin{align} {dy \over dx} & = 8x - 27x^{-2} \\ \\ {d^2 y \over dx^2} & = 8(1) - 27(-2)x^{-3} \\ & = 8 + 54 x^{-3} \\ & = 8 + {54 \over x^3} \\ \\ \text{When } & x = {3 \over 2}, \\ {d^2y \over dx^2} & = 8 + {54 \over ({3 \over 2})^3} \\ & = 24 \\ \\ \therefore \left({3 \over 2}, 27 \right) & \text{ is a minimum point} \end{align}

\begin{align} p^2q & = 9 \\ q & = {9 \over p^2} \\ \\ z & = 16p + 3q \\ \\ \text{Since } & q = {9 \over p^2}, \\ z & = 16p + 3\left(9 \over p^2\right) \\ & = 16p + {27 \over p^2} \\ & = 16p + 27p^{-2} \\ \\ {dz \over dp} & = 16(1) + 27(-2)(p^{-3}) \\ & = 16 - 54p^{-3} \\ & = 16 - {54 \over p^3} \\ \\ {d^2 z \over dp^2} & = -54(-3)p^{-4} \\ & = 162p^{-4} \\ & = {162 \over p^4} \\ \\ \text{Let } & {dz \over dp} = 0, \\ 0 & = 16 - {54 \over p^3} \\ {54 \over p^3} & = 16 \\ 54 & = 16p^3 \\ {54 \over 16} & = p^3 \\ {27 \over 8} & = p^3 \\ \sqrt[3]{27 \over 8} & = p \\ {3 \over 2} & = p \\ \\ \text{Substitute } & p = {3 \over 2} \text{ into } q = {9 \over p^2}, \\ q & = {9 \over ({3 \over 2})^2} \\ & = 4 \\ \\ \text{Substitute } & p = {3 \over 2} \text{ into } {d^2 z \over dp^2}, \\ {d^2z \over dp^2} & = {162 \over ({3 \over 2})^4} \\ & = 32 \\ \\ \therefore \text{When } p = {3 \over 2} & \text{ and } q = 4, \phantom{.} z \text{ is a minimum} \end{align}

(i)

\begin{align} \text{Total length of edges } & = 4 \times 3x + 4 \times x + 4 \times h \\ 2400 & = 12x + 4x + 4h \\ 2400 - 16x & = 4h \\ 600 - 4x & = h \\ \\ h & = 600 - 4x \end{align}

(ii)

\begin{align} V & = \text{Length} \times \text{Breadth} \times \text{Height} \\ & = 3x \times x \times h \\ & = 3x^2h \\ \\ \text{Since } & h = 600 - 4x, \\ V & = 3x^2(600 - 4x) \\ & = 3x^2(4)(150 - x) \\ & = 12x^2(150 - x) \end{align}

(iii)

\begin{align} V & = 12x^2 (150 - x) \\ & = 1800x^2 - 12x^3 \\ \\ {dV \over dx} & = 1800(2)(x) - 12(3)(x^2) \\ & = 3600x - 36x^2 \\ \\ {d^2 V \over dx^2} & = 3600(1) - 36(2)x \\ & = 3600 - 72x \\ \\ \text{Let } & {dV \over dx} = 0, \\ 0 & = 3600x - 36x^2 \\ 0 & = 36x(100 - x) \\ \\ 36x = 0 \phantom{0000(} & \text{or} \phantom{000} 100 - x = 0 \\ x = 0 \text{ (Rej)} & \phantom{or000100} - x = -100 \\ & \phantom{or0001000)(} x = 100 \\ \\ \text{Substitute } & x = 100 \text{ into } {d^2 V \over dx^2}, \\ {d^2V \over dx^2} & = 3600 - 72(100) \\ & = -3600 \\ \\ \therefore \text{When } & x = 100, \text{ the volume is a maximum} \end{align}

(i)

\begin{align} \text{Length of wire } & = \text{Perimeter of 1st square } + \text{Perimeter of 2nd square} \\ 100 & = 4x + 4y \\ 25 & = x + y \\ 25 - x & = y \\ \\ y & = 25 - x \end{align}

(ii)

\begin{align} A & = \text{Area of 1st square } + \text{Area of 2nd square} \\ & = (x)(x) + (y)(y) \\ & = x^2 + y^2 \\ \\ \text{Since } & y = 25 - x, \\ A & = x^2 + (25 - x)^2 \phantom{000} \text{ (Shown)} \end{align}

(iii)

\begin{align} A & = x^2 + (25 - x)^2 \\ \\ {dA \over dx} & = 2x + 2(25 - x). (-1) \phantom{00000} [\text{Chain rule}] \\ & = 2x - 2(25 - x) \\ & = 2x - 50 + 2x \\ & = 4x - 50 \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 4x - 50 \\ 50 & = 4x \\ {50 \over 4} & = x \\ 12.5 & = x \\ \\ \\ {dA \over dx} & = 4x - 50 \\ \\ {d^2 A \over dx^2} & = 4(1) \\ & = 4 \\ \\ \therefore \text{When } & x = 12.5, \phantom{.} A \text{ is a minimum} \end{align}

(i)

\begin{align} \text{Let } h \text{ rep} & \text{resent the height of the box} \\ \\ \text{Volume } & = \text{Length } \times \text{Breadth } \times \text{Height} \\ 100 & = 2x \times x \times h \\ 100 & = 2x^2h \\ {100 \over 2x^2} & = h \\ {50 \over x^2} & = h \\ \\ h & = {50 \over x^2} \\ \\ \\ \text{Total surface area, }S & = 2 \times (x)(2x) + 2 \times (x)(h) + 2 \times (2x)(h) \\ & = 4x^2 + 2xh + 4xh \\ & = 4x^2 + 6xh \\ \\ \text{Since } & h = {50 \over x^2}, \\ S & = 4x^2 + 6x\left(50 \over x^2\right) \\ & = 4x^2 + {300x \over x^2} \\ & = 4x^2 + {300 \over x} \phantom{000} \text{ (Shown)} \end{align}

(ii)

\begin{align} S & = 4x^2 + {300 \over x} \\ & = 4x^2 + 300x^{-1} \\ \\ {dS \over dx} & = 4(2)x + 300(-1)x^{-2} \\ & = 8x - 300x^{-2} \\ & = 8x - {300 \over x^2} \\ \\ \text{Let } & {dS \over dx} = 0, \\ 0 & = 8x - {300 \over x^2} \\ {300 \over x^2} & = 8x \\ 300 & = 8x^3 \\ {300 \over 8} & = x^3 \\ 37.5 & = x^3 \\ \\ x^3 & = 37.5 \\ x & = \sqrt[3]{37.5} \\ & = 3.34716 \\ \\ \text{Substitute } & x = 3.34716 \text{ into } S = 4x^2 + {300 \over x}, \\ S & = 4(3.34716)^2 + {300 \over (3.34716)} \\ & = 134.44214 \\ & \approx 134 \end{align}

(iii)

\begin{align} {dS \over dx} & = 8x - 300x^{-2} \\ \\ {d^2 S \over dx^2} & = 8(1) - 300(-2)x^{-3} \\ & = 8 + 600x^{-3} \\ & = 8 + {600 \over x^3} \\ \\ \text{When } & x = 3.34716, \\ {d^2S \over dx^2} & = 8 + {600 \over (3.34716)^3} \\ & = 24 \\ \\ \therefore \phantom{.} & S = 134 \text{ is a minimum} \end{align}

(i)

\begin{align} \text{Perimeter of figure } & = x + x + \text{Circumference of semi-circle } \\ 80 & = 2x + \pi r \\ 80 - \pi r & = 2x \\ \\ 2x & = 80 - \pi r \\ x & = {1 \over 2}(80 - \pi r) \\ \\ A & = \text{Area of right-angled triangle } + \text{Area of semi-circle} \\ & = {1 \over 2} \times \text{Base} \times \text{Height} + {1 \over 2} \pi r^2 \\ & = {1 \over 2}(x)(x) + {\pi r^2 \over 2} \\ & = {1 \over 2}x^2 + {\pi r^2 \over 2} \\ \\ \text{Since } & x = {1 \over 2}(80 - \pi r), \\ A & = {1 \over 2}\left[ {1 \over 2}(80 -\pi r) \right]^2 + {\pi r^2 \over 2} \\ & = {1 \over 2}\left(1 \over 2\right)^2(80 - \pi r)^2 + {\pi r^2 \over 2} \\ & = {1 \over 2}\left(1 \over 4\right)(80 - \pi r)^2 + {\pi r^2 \over 2} \\ & = {1 \over 8}(80 - \pi r)^2 + {\pi r^2 \over 2} \phantom{000} \text{ (Shown)} \end{align}

(ii)

\begin{align} A & = {1 \over 8}(80 - \pi r)^2 + {\pi r^2 \over 2} \\ \\ {dA \over dr} & = {1 \over 8}(2)(80 - \pi r).(-\pi) + {\pi (2)r \over 2} \\ & = -{1 \over 4}\pi (80 - \pi r) + \pi r \\ & = -20\pi + {1 \over 4} \pi^2 r + \pi r \\ \\ \text{Let } & {dA \over dr} = 0, \\ 0 & = -20\pi + {1 \over 4} \pi^2 r + \pi r \\ 0 & = {1 \over 4}\pi (-80 + \pi r + 4r) \\ 0 & = -80 + \pi r + 4r \\ -\pi r - 4r & = -80 \\ \pi r + 4r & = 80 \\ r(\pi + 4) & = 80 \\ r & = {80 \over \pi + 4} \\ & = 11.201 \\ & \approx 11.2 \end{align}

(iii)

\begin{align} {dA \over dr} & = -20\pi + {1 \over 4} \pi^2 r + \pi r \\ \\ {d^2 A \over dr^2} & = {1 \over 4} \pi^2 (1) + \pi (1) \\ & = {1 \over 4} \pi^2 + \pi \\ \\ \therefore \text{When } & r \approx 11.2, \phantom{.} A \text{ is a minimum} \end{align}

(i)

\begin{align} y & = x^2 + {16 \over x^2} \\ & = x^2 + 16x^{-2} \\ \\ {dy \over dx} & = (2)(x) + 16(-2)(x^{-3}) \\ & = 2x - 32x^{-3} \\ & = 2x - {32 \over x^3} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 2x - {32 \over x^3} \\ {32 \over x^3} & = 2x \\ 32 & = 2x^4 \\ 16 & = x^4 \\ \pm \sqrt[4]{16} & = x \\ \pm 2 & = x \end{align} \begin{align} x & = 2 && \text{ or } & x & = -2 \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = (2)^2 + {16 \over (2)^2} &&& y & = (-2)^2 + {16 \over (-2)^2} \\ & = 8 &&& & = 8 \\ \\ \therefore & \phantom{.} (2, 8) &&& \therefore & \phantom{.} (-2, 8) \end{align}

(ii)

\begin{align} {dy \over dx} & = 2x - 32x^{-3} \\ \\ {d^2 y \over dx^2} & = 2(1) - 32(-3)x^{-4} \\ & = 2 + 96x^{-4} \\ & = 2 + {96 \over x^4} \end{align} \begin{align} \text{When } & x = 2, &&& \text{When } & x = -2, \\ {d^2y \over dx^2} & = 2 + {96 \over (2)^4} &&& {d^2y \over dx^2} & = 2 + {96 \over (-2)^4} \\ & = 8 &&& & = 8 \\ \\ \therefore (2, 8) & \text{ is a min. point} &&& \therefore (-2, 8) & \text{ is a min. point} \end{align}

(i)

\begin{align} y & = {4 \over 2 - x} + {9 \over x - 3} \\ & = 4(2 - x)^{-1} + 9(x - 3)^{-1} \\ \\ {dy \over dx} & = 4(-1)(2 - x)^{-2}.(-1) + 9(-1)(x - 3)^{-2}.(1) \phantom{00000} [\text{Chain rule}] \\ & = 4(2 - x)^{-2} - 9(x - 3)^{-2} \\ & = {4 \over (2 - x)^2} - {9 \over (x - 3)^2} \\ \\ {d^2 y \over dx^2} & = 4(-2)(2 - x)^{-3}.(-1) - 9(-2)(x - 3)^{-3} . (1) \phantom{00000} [\text{Chain rule}] \\ & = 8(2 - x)^{-3} + 18 (x - 3)^{-3} \\ & = {8 \over (2 - x)^3} + {18 \over (x - 3)^3} \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = {4 \over (2 - x)^2} - {9 \over (x - 3)^2} \\ {9 \over (x - 3)^2} & = {4 \over (2 - x)^2} \\ 9(2 - x)^2 & = 4(x - 3)^2 \\ 9(4 - 4x + x^2) & = 4(x^2 - 6x + 9) \\ 36 - 36x + 9x^2 & = 4x^2 - 24x + 36 \\ 5x^2 - 12x & = 0 \\ x(5x - 12) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & 5x - 12 & = 0 \\ & &&& 5x & = 12 \\ & &&& x & = {12 \over 5} \\ & &&& & = 2.4 \\ \\ \text{Substitute} & \text{ into eqn for } y, &&& \text{Substitute} & \text{ into eqn for } y, \\ y & = {4 \over 2 - (0)} + {9 \over (0) - 3} &&& y & = {4 \over 2 - (2.4)} + {9 \over (2.4) - 3} \\ & = -1 &&& & = -25 \\ \\ \text{Substitute } & x = 0 \text{ into } {d^2y \over dx^2}, &&& \text{Substitute } & x = 2.4 \text{ into } {d^2y \over dx^2}, \\ {d^2y \over dx^2} & = {8 \over [2 - (0)]^3} + {18 \over [(0) - 3]^3} &&& {d^2y \over dx^2} & = {8 \over [2 - (2.4)]^3} + {18 \over [(2.4) - 3]^3} \\ & = {1 \over 3} &&& & = -208{1 \over 3} \\ \\ \therefore y = -1 & \text{ is a minimum value} &&& \therefore y = -25 & \text{ is a maximum value} \end{align}

\begin{align} \text{Let } x \text{ denote } & \text{the length of the rectangle} \\ \text{Let } y \text{ denote } & \text{the breadth of the rectangle} \\ \\ \text{Length of wire } & = \text{Perimeter of rectangle} \\ l & = 2x + 2y \\ l - 2x & = 2y \\ {1 \over 2}l - x & = y \\ \\ y & = {1 \over 2}l - x \\ \\ \\ \text{Let } A \text{ denote } & \text{the area of the rectangle} \\ \\ A & = \text{Length} \times \text{Breadth} \\ & = x \times y \\ & = xy \\ \\ \text{Since } & y = {1 \over 2}l - x, \\ A & = x\left({1 \over 2}l - x\right) \\ & = {1 \over 2}l x - x^2 \\ \\ {dA \over dx} & = {1 \over 2}l (1) - 2x \phantom{00000} [l \text{ is a constant}] \\ & = {1 \over 2}l - 2x \\ \\ {d^2 A \over dx^2} & = -2(1) \\ & = -2 \implies A \text{ is a maximum} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = {1 \over 2}l - 2x \\ 2x & = {1 \over 2}l \\ x & = {1 \over 4}l \\ \\ \text{Substitute } & x = {1 \over 4}l \text{ into } y = {1 \over 2}l - x, \\ y & = {1 \over 2}l - {1 \over 4}l \\ & = {1 \over 4} l \\ \\ \therefore x & = y = {1 \over 4}l \\ \\ \therefore A \text{ is a maximum} & \text{ when it is a square with length } {1 \over 4}l \text{ cm} \end{align}

(i) Since AB = AC, triangle ABC is an isosceles triangle.

Since R is the midpoint of BC, BR = RC = 5 cm and AR is the height of triangle ABC.

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ AB^2 & = AR^2 + BR^2 \\ 13^2 & = AR^2 + 5^2 \\ 169 & = AR^2 + 25 \\ 169 - 25 & = AR^2 \\ 144 & = AR^2 \\ \\ AR^2 & = 144 \\ AR & = \pm \sqrt{144} \\ AR & = \pm 12 \\ \\ AR & = 12 \text{ or } -12 \text{ (N.A.)} \\ \\ \\ \text{Comparing } & \text{similar triangles } APQ \text{ and } ABC, \\ {PQ \over BC} & = {\text{Height of }\triangle APQ \over AR} \\ {x \over 10} & = {12 - h \over 12} \\ 12x & = 10(12 - h) \\ 12x & = 120 - 10h \\ 10h & = 120 - 12x \\ h & = 12 - {6 \over 5}x \\ h & = {6 \over 5}(10 - x) \end{align}

(ii)

\begin{align} A & = {1 \over 2} \times \text{Base} \times \text{Height} \\ & = {1 \over 2}(x)(h) \\ & = {1 \over 2}xh \\ \\ \text{Since } & h = {6 \over 5}(10 - x), \\ A & = {1 \over 2}x\left[ {6 \over 5}(10 - x) \right] \\ & = {3 \over 5}x (10 - x) \\ & = {3x \over 5}(10 - x) \phantom{000} \text{ (Shown)} \end{align}

(iii)

\begin{align} A & = {3x \over 5}(10 - x) \\ & = 6x - {3x^2 \over 5} \\ \\ {dA \over dx} & = 6(1) - {3(2)(x) \over 5} \\ & = 6 - {6x \over 5} \\ \\ {d^2 A \over dx^2} & = - {6(1) \over 5} \\ & = -{6 \over 5} \implies A \text{ is a maximum} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 6 - {6x \over 5} \\ {6x \over 5} & = 6 \\ 6x & = 5(6) \\ 6x & = 30 \\ x & = {30 \over 6} \\ x & = 5 \\ \\ \therefore \text{When } & x = 5, \phantom{.} A \text{ is a maximum} \end{align}

(i)

\begin{align} \text{By comparing} & \text{ similar triangles } ABC \text{ and } AYP, \\ {AB \over AY} & = {BC \over YP} \\ {3 \over 3 - y} & = {4 \over x} \\ 3x & = 4(3 - y) \\ 3x & = 12 - 4y \\ 4y & = 12 - 3x \\ y & = 3 - {3 \over 4}x \\ y & = {3 \over 4}(4 - x) \end{align}

(ii)

\begin{align} A & = BX \times YB \\ & = x \times y \\ & = xy \\ \\ \text{Since } & y = {3 \over 4}(4 - x), \\ A & = x\left[{3 \over 4}(4 - x)\right] \\ & = {3 \over 4}x(4 - x) \phantom{000} \text{ (Shown)} \end{align}

(iii)

\begin{align} A & = {3 \over 4}x(4 - x) \\ & = 3x - {3 \over 4}x^2 \\ \\ {dA \over dx} & = 3(1) - {3 \over 4}(2)(x) \\ & = 3 - {3 \over 2}x \\ \\ {d^2 A \over dx^2} & = - {3 \over 2}(1) \\ & = -{3 \over 2} \implies A \text{ is a maximum} \\ \\ \text{Let } & {dA \over dx} = 0, \\ 0 & = 3 - {3 \over 2}x \\ {3 \over 2}x & = 3 \\ 3x & = 2(3) \\ 3x & = 6 \\ x & = {6 \over 3} \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into } A = {3 \over 4}x(4 - x), \\ A & = {3 \over 4}(2)[4 - (2)] \\ & = 3 \\ \\ \therefore \text{Maximum} & \text{ area} = 3 \text{ cm}^2 \end{align}

(i) Note: This triangle is an isosceles triangle

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ (6 + x)^2 & = h^2 + (8 - x)^2 \\ 36 + 12x + x^2 & = h^2 + 64 - 16x + x^2 \\ 36 - 64 + 12x +16x + x^2 - x^2 & = h^2 \\ -28 + 28x & = h^2 \end{align} \begin{align} h^2 & = 28x - 28 \\ h^2 & = 28(x - 1) \\ h & = \pm \sqrt{28(x - 1)} \\ \\ h & = \sqrt{28(x - 1)} \text{ or } -\sqrt{28(x - 1)} \text{ (N.A.)} \\ \\ A & = {1 \over 2} \times \text{Base} \times \text{Height} \\ & = {1 \over 2} \times (16 - 2x) \times \sqrt{28(x - 1)} \\ & = (8 - 2x)\sqrt{28(x - 1)} \phantom{000} \text{ (Shown)} \end{align}

(ii)

(i)

\begin{align} \text{Volume } & = \pi r^2 h \\ 250\pi & = \pi r^2 h \\ 250 & = r^2h \\ {250 \over r^2} & = h \\ \\ h & = {250 \over r^2} \\ \\ \\ A & = 2 \times \text{Base area} + \text{Curved surface area} \\ & = 2 \times \pi r^2 + 2\pi r h \\ & = 2\pi r^2 + 2\pi rh \\ \\ \text{Since } & h = {250 \over r^2}, \\ A & = 2\pi r^2 + 2\pi r\left( 250 \over r^2 \right) \\ & = 2\pi r^2 + {500\pi r \over r^2} \\ & = 2\pi r^2 + {500\pi \over r} \phantom{000} \text{ (Shown)} \end{align}

(ii) (a)

\begin{align} A & = 2\pi r^2 + {500\pi \over r} \\ & = 2\pi r^2 + 500\pi r^{-1} \\ \\ {dA \over dr} & = 2 \pi (2) r + 500 \pi (-1) r^{-2} \\ & = 4 \pi r - 500 \pi r^{-2} \\ & = 4 \pi r - {500 \pi \over r^2} \\ \\ {d^2 A \over dr^2} & = 4 \pi (1) - 500 \pi (-2) r^{-3} \\ & = 4\pi + 1000 \pi r^{-3} \\ & = 4\pi + {1000 \pi \over r^3} \\ \\ \text{Let } & {dA \over dr} = 0, \\ 0 & = 4\pi r - {500 \pi \over r^2} \\ {500 \pi \over r^2} & = 4\pi r \\ 500\pi & = 4\pi r^3 \\ {500 \pi \over 4\pi} & = r^3 \\ 125 & = r^3 \\ \sqrt[3]{125} & = r \\ 5 & = r \\ \\ \text{Substitute } & r = 5 \text{ into } h = {250 \over r^2}, \\ h & = {250 \over (5)^2} \\ & = 10 \\ \\ \text{Substitute } & r = 5 \text{ into } {d^2 A \over dr^2}, \\ {d^2A \over dr^2} & = 4\pi + {1000\pi \over (5)^3} \\ & = 12\pi \\ \\ \therefore \text{When } r = 5 \text{ and } & h = 10, \text{ area of material used is the least} \end{align}

(ii) (b)

\begin{align} \text{Let } & C \text{ denote the cost of material used} \\ \\ C & = \text{Base area} \times 0.03 + \text{Curved surface area} \times 0.05 \\ & = (2\pi r^2)\times 0.03 + \left(500\pi \over r\right)\times 0.05 \\ & = 0.06\pi r^2 + {25\pi \over r} \\ & = 0.06\pi r^2 + 25\pi r^{-1} \\ \\ {dC \over dr} & = 0.06\pi(2)r + 25\pi (-1)r^{-2} \\ & = 0.12\pi r - 25\pi r^{-2} \\ & = 0.12\pi r - {25\pi \over r^2} \\ \\ {d^2 C \over dr^2} & = 0.12\pi (1) - 25 \pi (-2) r^{-3} \\ & = 0.12\pi + 50 \pi r^{-3} \\ & = 0.12\pi + {50 \pi \over r^3} \\ \\ \text{Let } & {dC \over dr} = 0, \\ 0 & = 0.12\pi r - {25\pi \over r^2} \\ {25\pi \over r^2} & = 0.12\pi r \\ 25\pi & = 0.12\pi r^3 \\ {25\pi \over 0.12\pi} & = r^3 \\ 208{1 \over 3} & = r^3 \\ \\ r^3 & = 208{1 \over 3} \\ r & = \sqrt[3]{208{1 \over 3}} \\ & = 5.9281 \\ \\ \text{Substitute } & r = 5.9281 \text{ into } h = {250 \over r^2}, \\ h & = {250 \over (5.9281)^2} \\ & = 7.1139 \\ \\ \text{Substitute } & r = 5.9281 \text{ into } {d^2 C \over dr^2}, \\ {d^2C \over dr^2} & = 0.12\pi + {50\pi \over (5.9281)^3} \\ & = 1.1309 \\ \\ \therefore \text{When } r \approx 5.93 & \text{ and } h \approx 7.11, \text{ cost of material used is the least} \end{align}