Q. 26 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

The maximum area that can be enclosed is 25 square units. 

Step 1. Given information.

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

Step 2. Draw the figure.

The below figure gives us the better illustration of the length of the side of the rectangle.

Step 3. Find the area of rectangle.

Using the formula for the area of rectangle is given by: 

$A=lw$

We can obtain an equation for the area of rectangle given by:

$$\begin{gathered} A=\left(x\right)\left(10-x\right) \\ A=-x^2+10x \end{gathered}$$

Notice that the quadratic equation has downward opening parabola, thus the vertex of the equation will gives us the maximum area of the rectangle .

Let make use of the formula below to obtain the x-coordinates of the vertex:

$x=-\frac{b}{2a}$

where a and b are the coefficient of the equation in the form $a{x}^2+bx+x=0$.

Step 3. Substitute the values.

Substitute $a=-1 \mathrm{and} b=10$to the equation above to obtain the x-coordinate of the vertex:

$x=-\frac{10}{2\left(-1\right)}=5$

So, when $x=5$ the maximum area is obtained.

Now, substitute $x=5$ to the equation for area to get the maximum area that can be enclosed:

$$\begin{gathered} A=-x^2+10x \\ A=-{\left(5\right)}^2+10\left(5\right) \\ A=-25+50 \\ A=25 {\mathrm{unit}}^2 \end{gathered}$$

Therefore, the maximum area that can be enclosed is 25 square units.