Optimisation: Area Problems
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
A rectangle has perimeter 60 cm. Find the dimensions that maximise the area, and state the maximum area.
Let width \(=x\), length \(=30-x\)
\(A=x(30-x)=30x-x^2\)
Maximise
\(A'=30-2x=0\Rightarrow x=15\)
Answer
Square: \(15\times15\) cm; max area \(=225\) cm²
Question 2
A farmer encloses a rectangular area using a wall on one side and 80 m of fencing for the other three sides. Find the dimensions that maximise the enclosed area.
Let width \(=x\); length \(=80-2x\)
\(A=x(80-2x)\Rightarrow A'=80-4x=0\Rightarrow x=20\)
Answer
\(20\times40\) m; max area \(=800\) m²
Question 3
A triangle has a base on the \(x\)-axis with vertices at \((-a,0)\), \((a,0)\) and a third vertex directly above \(x=a\), on the curve \(y=9-x^2\). Find the dimensions of the triangle with maximum area.
Area in terms of \(a\)
\(A=\tfrac{1}{2}\times2a\times(9-a^2)=a(9-a^2)\)
Maximise
\(A'=9-3a^2=0\Rightarrow a=\sqrt{3}\)
Answer
Base \(=2\sqrt{3}\), height \(=6\); max area \(=6\sqrt{3}\) sq units
Question 4
A window consists of a rectangle surmounted by a semicircle. The total perimeter of the window is 10 m. Find the dimensions that maximise the area of the window.
Let rectangle width \(=2r\), height \(=h\); semicircle radius \(=r\)
Perimeter: \(2h+2r+\pi r=10\Rightarrow h=\tfrac{10-r(2+\pi)}{2}\)
Area
\(A=2rh+\tfrac{\pi r^2}{2}=r(10-r(2+\pi))+\tfrac{\pi r^2}{2}=10r-r^2(2+\pi)+\tfrac{\pi r^2}{2}=10r-r^2(2+\tfrac{\pi}{2})\)
Maximise
\(A'=10-r(4+\pi)=0\Rightarrow r=\tfrac{10}{4+\pi}\)
Answer
\(r=\dfrac{10}{4+\pi}\approx1.40\) m; \(h=\dfrac{10}{4+\pi}\approx1.40\) m (i.e. \(h=r\))
Question 5
Find the dimensions of the rectangle of maximum area that can be inscribed in the ellipse \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\).
Parametrise
\(x=4\cos\theta,\;y=3\sin\theta\); area \(=4xy=48\sin\theta\cos\theta=24\sin2\theta\)
Maximise
Max when \(\sin2\theta=1\Rightarrow\theta=\tfrac{\pi}{4}\Rightarrow x=\tfrac{4}{\sqrt2},\;y=\tfrac{3}{\sqrt2}\)
Answer
Dimensions \(2x\times2y=4\sqrt{2}\times3\sqrt{2}\) (\(\approx5.66\times4.24\)); max area \(=24\sin2\theta=24\) sq units
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