(a)(i)

Let $AB=y=\frac{40-x}{2}$

$h = \sqrt{y^2-(\frac{x}{2})^2}=2\sqrt{100-5x}$

$z = \frac{1}{2}xh=\frac{1}{2}x(2\sqrt{100-5x}=x\sqrt{100-5x}$

(a)(ii)

$\frac{dz}{dx}=\sqrt{100-5x} + x\frac{1}{2}(-5)(100-5x)^{-\frac{1}{2}}$

$\frac{dz}{dx}=\frac{100-\frac{15}{2}x}{\sqrt{100-5x}}$

$\frac{dz}{dx}=0 \Rightarrow x=\frac{40}{3}$

*Students are expected to prove that $x = \frac{40}{3}$ gives the maximum area.

(b)(i)
$x^3 + 2y^3 +3xy=k$

$3x^2+6y^2\frac{dy}{dx}+3x\frac{dy}{dx}+3y=0$

$\frac{dy}{dx}=\frac{-y-x^2}{2y^2+x} (b)(ii) Tangent parallel to$latex x\$-axis $\Rightarrow \frac{dy}{dx}=0$

$y=-x^2$

Sub $y=-x^2$ into $x^3 + 2y^3 +3xy=k$

$x^3 + 2(-x^2)^3 +3x(-x^2)=k$

$2x^6 + 2x^3 + k=0$

(b)(iii)
When the line $y=-1$ is a tangent to C,

$-1=-x^2$

$x =\pm x$

When $x = 1, k=-4$

When $x = -1, k=0$

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