Question #276101

Two gardens. A fence of length 100 ft is to be used to enclose two


gardens. One garden is to have a circular shape, and the other to be


square.


Determine how the fence should be cut so that the sum of the areas


inside both gardens is as large as possible.

Expert's answer

Let side of the square garden is a and radius of the circular garden is r.

We have to maximize a2+πr2a^{2}+\pi r^{2}

  Subject to, 4a+2πr=100 i.e., 2a+πr=50\begin{aligned} &\text { Subject to, } 4 a+2 \pi r=100 \\ &\text { i.e., } 2 a+\pi r=50 \end{aligned}

The Lagrange function L=(a2+πr2)+λ(2a+πr50).L=\left(a^{2}+\pi r^{2}\right)+\lambda(2 a+\pi r-50).

La2a+2λ=0λ=aLr2πr+λπ=0λ=2r.\therefore \begin{aligned} \frac{\partial L}{\partial a} & \equiv 2 a+2 \lambda=0 \quad \Rightarrow \quad \lambda=-a \\ \frac{\partial L}{\partial r} & \equiv 2 \pi r+\lambda \pi=0 \quad \Rightarrow \lambda=-2 r . \end{aligned}

2(λ)+π(λ2)=50λ(2π2)=50λ=1004+πa=1004+π and, r=504+π\begin{aligned} \therefore & 2(-\lambda)+\pi\left(-\frac{\lambda}{2}\right)=50 \\ & \Rightarrow \lambda\left(-2-\frac{\pi}{2}\right)=50 \\ & \Rightarrow \lambda=\frac{-100}{4+\pi} \\ \therefore \quad & a=\frac{100}{4+\pi} \\ \text { and, } & r=\frac{50}{4+\pi} \end{aligned}

\therefore The required fence for square =4004+πft=\frac{400}{4+\pi} \mathrm{ft} and for circular garden =100π4+πft.=\frac{100 \pi}{4+\pi} f_{t}.



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