Chapter 5.4, Problem 2E

To determine

Find the largest possible area of the triangle.

Answer to Problem 2E

The maximum area the sides are $\left(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}}\right)$ cm and area is $\frac{25}{4}c{m}^2$ .

Explanation of Solution

Given information: Given a right triangle whose hypotenuse of the right triangle is $5\text{cm}$

Calculation:

Let the other sides are,

  $\begin{array}{l}x\text{ and }y0<x<5\\ x^2+y^2={\left(5\right)}^2\\ x^2+y^2=25\\ y=\pm \sqrt{25-x^2}\end{array}$

Let the area of the triangle be ,

  $\begin{array}{l}A=\frac{1}{2}\left(x\right)\left(y\right)\\ =\frac{1}{2}\left(x\right)\left(\sqrt{25-x^2}\right)\end{array}$

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

  $\begin{array}{l}A=\frac{1}{2}\left(x\right)\left(\sqrt{25-x^2}\right)\\ \frac{dA}{dx}=\frac{1}{2}\left(\sqrt{25-x^2}\right)+\frac{1}{2}\left(x\right)\left(\frac{1\left(-2x\right)}{2\sqrt{25-x^2}}\right)\\ =\frac{25-2x2}{2\sqrt{25-x^2}}\end{array}$

Critical points occur at

  $\begin{array}{l}{A}^{\prime }=0\\ 25-2x^2=0\\ x=\pm \frac{5}{\sqrt{2}}\end{array}$

Consider $x=\frac{5}{\sqrt{2}}\text{ since }x>0$ ,solve it graphically, it has critical point occur at $x=\frac{5}{\sqrt{2}}$

  $\frac{dA}{dx}>0\text{ for }0<x<\frac{5}{\sqrt{2}}\text{and}\frac{dA}{dx}<0\text{for}\frac{5}{\sqrt{2}}<x<5$ this corresponds to maximum area.

For

  $\begin{array}{l}x=\frac{5}{\sqrt{2}}\\ y=\sqrt{25-x^2}\\ =\sqrt{25-{\left(\frac{5}{\sqrt{2}}\right)}^2}\\ =\frac{5}{\sqrt{2}}\\ A=\frac{1}{2}\left(x\right)\left(y\right)\\ =\frac{1}{2}\left(\frac{5}{\sqrt{2}}\right)\left(\frac{5}{\sqrt{2}}\right)\\ =\frac{25}{4}\end{array}$

Thus, the maximum area the sides are $\left(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}}\right)$ cm and area is $\frac{25}{4}c{m}^2$

Graphical support: