Optimisation (Mixed)
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
A farmer has 120 m of fencing and wishes to enclose a rectangular field against a straight wall (wall forms one side — no fencing needed on that side). Find the dimensions that maximise the area.
Setup
Let width \(=x\); then \(2x+\text{length}=120\Rightarrow\text{length}=120-2x\). Area: \(A=x(120-2x)=120x-2x^2\)
Maximise
\(\dfrac{dA}{dx}=120-4x=0\Rightarrow x=30\)
Answer
Width 30 m, length 60 m; max area \(=1800\) m²
Question 2
A cylindrical can (with lid) must hold \(500\pi\) cm³. Find the dimensions that minimise the total surface area.
Setup
\(V=\pi r^2 h=500\pi\Rightarrow h=\dfrac{500}{r^2}\)
Surface area
\(A=2\pi r^2+2\pi rh=2\pi r^2+\dfrac{1000\pi}{r}\)
Minimise
\(\dfrac{dA}{dr}=4\pi r-\dfrac{1000\pi}{r^2}=0\Rightarrow r^3=250\Rightarrow r=\sqrt[3]{250}\)
Answer
\(r=\sqrt[3]{250}\approx6.30\) cm; \(h=\dfrac{500}{r^2}\approx12.6\) cm
Question 3
A rectangle is inscribed in a circle of radius 5. Find the dimensions that maximise the area of the rectangle.
Setup using angle
Half-sides: \(x=5\cos\theta,\;y=5\sin\theta\); Area \(=4xy=100\sin\theta\cos\theta=50\sin2\theta\)
Maximise
Max when \(\sin2\theta=1\Rightarrow\theta=\tfrac{\pi}{4}\Rightarrow x=y=\dfrac{5}{\sqrt{2}}\)
Answer
Square with side \(\dfrac{10}{\sqrt{2}}=5\sqrt{2}\) cm; max area \(=50\) cm²
Question 4
The profit from selling \(x\) units is \(P(x)=-2x^2+120x-800\). Find the number of units to maximise profit and state the maximum profit.
Maximise
\(P'(x)=-4x+120=0\Rightarrow x=30\)
Maximum profit
\(P(30)=-1800+3600-800\)
Answer
30 units; max profit \(=\$1000\)
Question 5
A piece of wire 80 cm long is cut into two pieces. One piece is bent into a square, the other into a circle. Find the lengths that minimise the combined area.
Setup
Let square have perimeter \(x\); circle perimeter \(80-x\). Area: \(A=\left(\tfrac{x}{4}\right)^2+\pi\left(\tfrac{80-x}{2\pi}\right)^2=\dfrac{x^2}{16}+\dfrac{(80-x)^2}{4\pi}\)
Minimise
\(\dfrac{dA}{dx}=\dfrac{x}{8}-\dfrac{80-x}{2\pi}=0\Rightarrow \pi x=4(80-x)\Rightarrow x(\pi+4)=320\Rightarrow x=\dfrac{320}{\pi+4}\)
Answer
Square perimeter \(=\dfrac{320}{\pi+4}\approx44.8\) cm; circle circumference \(\approx35.2\) cm
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