Chapter 3, Problem 26RE

26. Maximizing Area A rectangle has one vertex on the line $y = 10 - x$. $x > 0$, another at the origin, one on the positive $x\text{-axis}$, and one on the positive $y\text{-axis}$. Express the area A of the rectangle as a function of x. Find the largest area A that can he enclosed by the rectangle.

To determine

To calculate: Express the area, $A$ as a function of $x$ and then find the maximum area that can be enclosed by the rectangle.

Answer to Problem 26RE

The maximum area that can be enclosed by the given rectangle is 25 square units.

Explanation of Solution

Given:

The vertexes of the rectangle are on the positive $x\text{-axis}$, positive $y\text{-axis}$, the origin and on the line $y=10-x,x>0$

Formula used:

The area of a rectangle is $Area=length\times width$

For a quadratic function $f(x)=a{x}^2+bx+c$, the vertex $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$ is the maximum point if $a$ is negative and the vertex is the minimum point if $a$ is positive.

Calculation:

Here, from the given information, we can see that the length on the rectangle is $x$ and the width is $y$.

Thus, area is

$A=x\times y=x(10-x)$

$=10x-x^2$

The area is a quadratic function with negative $a$.

Thus, the maximum value is the vertex.

Therefore, we have

$x=\frac{-b}{2a}=\frac{-(10)}{2(-1)}=5$

The maximum area is

$A(5)=10(5)-{(5)}^2=25$

Thus, the maximum area that can be enclosed by the given rectangle is 25 square units.