2012 H2 Mathematics Paper 1 Question 8

Differentiation II: Maxima, Minima, Rates of Change

Answers

(iii)

Maximum point

Full solutions

(i)

1−dydx=2(x+y)(1+dydx)2(x+y)(1+dydx)+dydx=12(x+y)(1+dydx)+dydx+1=2(1+dydx)(2(x+y)+1)=21+dydx=22x+2y+1  ■\begin{gather*} 1 - \frac{\mathrm{d}y}{\mathrm{d}x} = 2(x+y)(1+\frac{\mathrm{d}y}{\mathrm{d}x}) \\ 2(x+y)(1+\frac{\mathrm{d}y}{\mathrm{d}x}) + \frac{\mathrm{d}y}{\mathrm{d}x} = 1 \\ 2(x+y)(1+\frac{\mathrm{d}y}{\mathrm{d}x}) + \frac{\mathrm{d}y}{\mathrm{d}x} + 1 = 2 \\ (1+\frac{\mathrm{d}y}{\mathrm{d}x}) \Big( 2(x+y) + 1\Big) = 2 \\ 1 + \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{2x+2y+1} \; \blacksquare \end{gather*}

(ii)

d2ydx2=−2(2x+2y+1)2(2+2dydx)=−4(2x+2y+1)2(1+dydx)=−(1+dydx)2(1+dydx)=−(1+dydx)3  ■\begin{align*} \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} &= \frac{-2}{(2x+2y+1)^2} \left( 2+2\frac{\mathrm{d}y}{\mathrm{d}x} \right) \\ &= - \frac{4}{(2x+2y+1)^2} \left( 1+\frac{\mathrm{d}y}{\mathrm{d}x} \right) \\ &= - \left( 1 + \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 \left( 1+\frac{\mathrm{d}y}{\mathrm{d}x} \right) \\ &= - \left( 1+\frac{\mathrm{d}y}{\mathrm{d}x} \right)^3 \; \blacksquare \end{align*}

(iii)

At the turning point, dydx=0{\frac{\mathrm{d}y}{\mathrm{d}x}=0}
d2ydx2=−(1+0)2=−1<0\begin{align*} \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} &= - \left( 1 + 0 \right)^2 \\ &= - 1 \\ &< 0 \end{align*}
Hence the turning point is a maximum point ■{\blacksquare}