Answer to Question #246504 in Calculus for sad

Question

The perimeter of an isosceles triangle is 28cm. What is the maximum area possible? 

Answer 1

Let $x$ denotes the length of a leg of a triangle. Since the perimeter is 28 and the triangle is isoscales, the length of the base is $28-x-x=28-2x.$ Let $h$ be the length of the altitude to the base. Since the triangle is isoscales, the altitude and the median to the base are coinside. By Pythagorean theorem, $h=\sqrt{x^2-(\frac{28-2x}2)^2}=\sqrt{x^2-(14-x)^2}=\sqrt{x^2-196 +28x-x^2}=\sqrt{28x-196}=2\sqrt{7x-49}.$

Then the area of triangle is $A(x)=\frac{1}2(28-2x)2\sqrt{7x-49}=2(14-x)\sqrt{7x-49}.$

It follows that $A'(x)=2(-\sqrt{7x-49}+(14-x)\frac{7}{2\sqrt{7x-49}}) =2\frac{-2(7x-49)+7(14-x)}{2\sqrt{7x-49}} =\frac{196-21x}{\sqrt{7x-49}}.$

If $A'(x)=0$ then $196-21x=0,$ and hence $x=\frac{196}{21}=\frac{28}3.$ Since $A'(x)>0$ for $x<\frac{28}3$ and $A'(x)<0$ for $x>\frac{28}3,$ we conclude that $x=\frac{28}3$ is the point of maximum.

Taking into account that $A(\frac{28}3)=2(14-\frac{28}3)\sqrt{7\frac{28}3-49}=\frac{28}3\sqrt{\frac{49}3}=\frac{196\sqrt{3}}{9},$ we conclude that $\frac{196\sqrt{3}}{9}$ is the maximum area possible.