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A pentagon is made by mounting an isosceles triangle on top of a rectangle. What dimensions minimize perimeter $P$ for a given area $K$. This was asked in a test today.

Let $x$ and $y$ denote the sides of the rectangle, and let $z$ be the height of the triangle mounted on the side with length $y$.

The perimeter is $P=2x+y+2\sqrt{0.25y^2 + z^2}$ while the area is $K = xy + 0.5yz$.

My attempt at solving it was equating dP/DK to zero. The resulting expression was extremely messy and very hard to rewrite. Is there a better, more elegant way to solve this than by forcing the answer through repeated application of chain rule.

Any answer would be appriciated.

LinAlg
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Gagan Surathkal
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  • I have added some formulas to your question to assist others in answering the question. – LinAlg Jan 18 '21 at 16:22
  • Using standard methods of multivariable calculus together with brute-force algebra, the result I get is that for a fixed area $K$, the minimum perimeter $P$ is equal to $\sqrt{K(8+4\sqrt{3})}$ which is achieved using $$ x=\sqrt{K\left(\frac{1}{2}-\frac{\sqrt{3}}{6}\right)}\qquad y=\sqrt{K\left(2-\sqrt{3}\right)}\qquad z=\sqrt{K\left(\frac{\sqrt{3}}{3}-\frac{1}{2}\right)} $$ – quasi Jan 18 '21 at 19:57

1 Answers1

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The optimization problem is: $$\min_{x,y,z} \left\{ 2x+y+2\sqrt{0.25y^2 + z^2} : xy + 0.5yz = K\right\}$$ The Lagrangian is: $$L(x,y,z,\lambda) =2x+y+2\sqrt{0.25y^2 + z^2} + \lambda(xy+0.5yz-K)$$ The first order conditions are: $$2+\lambda y =0 \quad (1)$$ $$1+\frac{0.5y}{\sqrt{0.25y^2+z^2}}+\lambda x + 0.5 \lambda z = 0 \quad (2)$$ $$\frac{2z}{\sqrt{0.25y^2+z^2}}+0.5 \lambda y = 0 \quad (3)$$ You can substitute (1) into (3) to get $z^2 = 4x^2 - 0.25y^2$. Substitute that into (2): $$1+\frac{0.5y}{2x}+\lambda x + 0.5 \lambda \sqrt{4x^2 - 0.25y^2} = 0 \quad$$

LinAlg
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