2019 H2 Mathematics Paper 2 Question 3

Differentiation II: Maxima, Minima, Rates of Change

Answers

Maximum possible volume =15006π cm3{ = 1500 \sqrt{\frac{6}{\pi}} \textrm{ cm}^3}
r:h=1:2{r:h = 1:2}

Full solutions

2πrh+2πr2=900h=1πr(450−πr2)\begin{gather*} 2\pi r h + 2 \pi r^2 = 900 \\ h = \frac{1}{\pi r} \left( 450 - \pi r^2 \right) \end{gather*}
Let V{V} denote the volume of the cylinder
V=πr2h=πr2(1πr(450−πr2))=450r−πr3\begin{align*} V &= \pi r^2 h \\ &= \pi r^2 \left( \frac{1}{\pi r} \left( 450 - \pi r^2 \right) \right) \\ &= 450r - \pi r^3 \end{align*}
dVdr=450−3πr2\frac{\mathrm{d}V}{\mathrm{d}r} = 450 - 3 \pi r^2
At stationary values of V,{V, } dVdr=0{\displaystyle \frac{\mathrm{d}V}{\mathrm{d}r} = 0}
450−3πr2=0r=150π\begin{gather*} 450 - 3 \pi r^2 = 0 \\ r = \sqrt{\frac{150}{\pi}} \end{gather*}
V=450150π−π(150π)3=150π(450−π(150π))=300150π=15006π cm3  ■\begin{align*} V &= 450 \sqrt{\frac{150}{\pi}} - \pi \left( \sqrt{\frac{150}{\pi}} \right)^3 \\ &= \sqrt{\frac{150}{\pi}} \left( 450 - \pi \left( \frac{150}{\pi} \right) \right) \\ &= 300 \sqrt{\frac{150}{\pi}} \\ &= 1500 \sqrt{\frac{6}{\pi}} \textrm{ cm}^3 \; \blacksquare \end{align*}
d2Vdr2=−6πr<0\begin{align*} \frac{\mathrm{d}^{2}V}{\mathrm{d}r^{2}} &= -6 \pi r \\ <0 \end{align*}
Hence V=15006π{V = 1500 \sqrt{\frac{6}{\pi}}} is a maximum
rh=r÷1πr(450−πr2)=πr2450−πr2=π150π450−π150π=150300=12\begin{align*} \frac{r}{h} &= r \div \frac{1}{\pi r} \left( 450 - \pi r^2 \right) \\ &= \frac{\pi r^2}{450-\pi r^2} \\ &= \frac{\pi \frac{150}{\pi}}{450 - \pi \frac{150}{\pi}} \\ &= \frac{150}{300} \\ &= \frac{1}{2} \end{align*}
r:h=1:2  ■r:h = 1:2 \; \blacksquare