Chapter 5.4, Problem 8E

To determine

Prove the maximum area of the dimensions are $a=b$

Answer to Problem 8E

The maximum area of the dimensions are $a=b\text{ is }\left(14.14,14.14\right)$ and area is $100\text{units}$

Explanation of Solution

Given information: Given the length of the line is $20$ units with coordinates $\left(a,0\right)\left(0,b\right)$

Calculation:

Let the sides length of the triangle is $a,b\left(0<a<20\right)$

  $\begin{array}{l}\left(a^2+b^2\right)={20}^2\\ b^2=400-a^2\\ b=\sqrt{400-a^2}\end{array}$

Let the area of the triangle be ,

  $\begin{array}{l}A=\frac{1}{2}\left(a\right)\left(b\right)\\ =\frac{1}{2}\left(a\right)\left(\sqrt{400-a^2}\right)\end{array}$

For any function $f\left(x\right)$ the condition of minimum and maximum is obtained by find $f^{\prime }\left(x\right)=0$ ,

Critical points and end points

  $\begin{array}{l}A=\frac{1}{2}\left(a\right)\left(\sqrt{400-a^2}\right)\\ \frac{dA}{da}=\frac{1}{2}\left(\sqrt{400-a^2}\right)+\frac{\left(a\right)}{2\left(2\sqrt{25-x^2}\right)}\left(-2a\right)\\ =\frac{1}{2}\left(\sqrt{400-a^2}\right)-\frac{\left(a^2\right)}{2\left(\sqrt{400-a^2}\right)}\\ =\frac{\left(400-2a^2\right)}{2\left(\sqrt{400-a^2}\right)}\\ =\frac{\left(200-a^2\right)}{\left(\sqrt{400-a^2}\right)}\end{array}$

Critical points occur at

  $\begin{array}{l}{A}^{\prime }=0\\ \frac{\left(200-a^2\right)}{\left(\sqrt{400-a^2}\right)}=0\\ 200-a^2=0\\ a=\pm \sqrt{200}\end{array}$

Consider $a=\sqrt{200}\text{ since }0<a<20$ ,solve it graphically, it has critical point occur at $a=\sqrt{200}$

Here $\frac{dA}{da}>0\text{ for }0<a<\sqrt{200}\text{ and }\frac{dA}{da}<0\text{for}\sqrt{200}<a<20$ this corresponds to maximum area.

For

  $\begin{array}{l}a=\sqrt{200}\\ b=\left(\sqrt{400-a^2}\right)\\ =\sqrt{400-200}\\ =\sqrt{200}\approx 14.14\\ a=b\\ A=\frac{1}{2}\left(\sqrt{200}\right)\left(\sqrt{200}\right)\\ =100\end{array}$

Graphical support:

  

Thus, the maximum area of the dimensionsare $a=b\text{ is }\left(14.14,14.14\right)$ and area is $100\text{units}$