Lexi Mcneil
2022-07-19
Answered

How do you find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle?

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Makenna Lin

Answered 2022-07-20
Author has **16** answers

Let the upper base y of the rectangle be the segment of a line parallel to the base of the equilateral triangle at an unknown distance x from it. In such a way the triangle is divided in two triangles, the equilateral one having height $h=L\frac{\sqrt[2]{3}}{2}$ and

a smaller one having height ${h}_{1}=L\frac{\sqrt[2]{3}}{2}-x$, that are similar! so we can write the

proportion $\frac{L}{y}=\frac{L\frac{\sqrt[2]{3}}{2}}{L\frac{\sqrt[2]{3}}{2}-x}$. By insulating the y we obtain $y=L-\frac{2}{\sqrt[2]{3}}x$

The rectangle area is $S(x,y)=x\cdot y$ but

$S\left(x\right)=x\cdot (L-\frac{2}{\sqrt[2]{3}}x)=Lx-\frac{2}{\sqrt[2]{3}}{x}^{2}$

By deriving S(x) we get $S\prime \left(x\right)=L-\frac{4}{\sqrt[2]{3}}x$ whose root is $x=L\frac{\sqrt[2]{3}}{4}$ and

consequently $y=L-\frac{2}{\sqrt[2]{3}}\frac{\sqrt[2]{3}}{4}L=\frac{L}{2}$

a smaller one having height ${h}_{1}=L\frac{\sqrt[2]{3}}{2}-x$, that are similar! so we can write the

proportion $\frac{L}{y}=\frac{L\frac{\sqrt[2]{3}}{2}}{L\frac{\sqrt[2]{3}}{2}-x}$. By insulating the y we obtain $y=L-\frac{2}{\sqrt[2]{3}}x$

The rectangle area is $S(x,y)=x\cdot y$ but

$S\left(x\right)=x\cdot (L-\frac{2}{\sqrt[2]{3}}x)=Lx-\frac{2}{\sqrt[2]{3}}{x}^{2}$

By deriving S(x) we get $S\prime \left(x\right)=L-\frac{4}{\sqrt[2]{3}}x$ whose root is $x=L\frac{\sqrt[2]{3}}{4}$ and

consequently $y=L-\frac{2}{\sqrt[2]{3}}\frac{\sqrt[2]{3}}{4}L=\frac{L}{2}$

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