Optimisation: Volume Problems
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
A open-topped box is made from a square sheet of card (side 24 cm) by cutting equal squares from each corner and folding up the sides. Find the size of corner square that maximises the volume.
Let corner square have side \(x\)
\(V=x(24-2x)^2\)
Expand and differentiate
\(V=x(576-96x+4x^2)=576x-96x^2+4x^3\)
\(V'=576-192x+12x^2=12(x^2-16x+48)=12(x-4)(x-12)=0\)
Reject \(x=12\) (no box left)
Answer
\(x=4\) cm; max volume \(=4 imes16^2=1024\) cm³
Question 2
An open cylindrical can must hold \(250\pi\) cm³. Find the radius that minimises the total surface area (no lid).
Volume
\(\pi r^2 h=250\pi\Rightarrow h= frac{250}{r^2}\)
Surface area (no lid)
\(A=\pi r^2+2\pi rh=\pi r^2+ frac{500\pi}{r}\)
Minimise
\(A'=2\pi r- frac{500\pi}{r^2}=0\Rightarrow r^3=250\Rightarrow r=\sqrt[3]{250}pprox6.30\)
Answer
\(r=\sqrt[3]{250}pprox6.30\) cm; \(hpprox6.30\) cm (cube-like shape)
Question 3
A cone has slant height 12 cm. Find the radius that maximises the volume of the cone.
Height in terms of \(r\)
\(h=\sqrt{144-r^2}\)
Volume
\(V= frac{\pi r^2}{3}\sqrt{144-r^2}\)
Maximise \(V^2\) (simpler)
\(V^2= frac{\pi^2}{9}r^4(144-r^2)\); differentiate: \( frac{d(V^2)}{dr}= frac{\pi^2}{9}(576r^3-6r^5)=0\Rightarrow r^2=96\Rightarrow r=4\sqrt6\)
Answer
\(r=4\sqrt{6}pprox9.80\) cm; \(h=\sqrt{144-96}=4\sqrt{3}pprox6.93\) cm
Question 4
A rectangular box with a square base and lid has volume 32 cm³. Find the dimensions that minimise the total surface area.
Let base side \(=x\), height \(=h\)
\(x^2h=32\Rightarrow h= frac{32}{x^2}\)
Surface area
\(A=2x^2+4xh=2x^2+ frac{128}{x}\)
Minimise
\(A'=4x- frac{128}{x^2}=0\Rightarrow x^3=32\Rightarrow x=2\sqrt[3]{4}\)
Answer
\(x=2\sqrt[3]{4}pprox3.17\) cm; \(h= frac{32}{x^2}pprox3.17\) cm (cube)
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