June Revision Exercise 3 Q3

By KS Teng

(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 tolatex x-axislatex \Rightarrow \frac{dy}{dx}=0latex y=-x^2Sublatex y=-x^2intolatex x^3 + 2y^3 +3xy=klatex x^3 + 2(-x^2)^3 +3x(-x^2)=klatex 2x^6 + 2x^3 + k=0(b)(iii) When the linelatex y=-1is a tangent to C,latex -1=-x^2latex x =\pm xWhenlatex x = 1, k=-4Whenlatex x = -1, k=0$

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