Question

Patio Design. A stone mason has enough stones to enclose a rectangular patio with $60$ ft of perimeter, assuming that the attached house forms one side of the rectangle. What is the maximum area that the mason can enclose?

What should the dimensions of the patio be in order to yield this area?

Answer 1

The maximum area that the mason can enclose:

The perimeter of the rectangular Patio is $60$ ft.

Since the attached house form one side of rectangle, the rectangular patio has three sides.

If L ft is length and W is width of the patio then the perimeter of the rectangular patio is $L+2W$ which is equal to $60$ ft.

That is,

$L+2W=60$.

The area of the patio is $L\times W$

From, $L+2W=60$, substitute $W=\frac{\left(60-L\right)}{2}$in $\text{Area}=L\times W$.

$$\begin{gathered} \text{Area}=L\times \frac{60-L}{2} \\ =\frac{1}{2}\times \left(60L-L^2\right) \end{gathered}$$

Now, to have maximum area differentiate the above equation with respect to L and equate it to zero:

$$\begin{gathered} \frac{1}{2}\times \frac{d\left(60L-L^2\right)}{dL}=0 \\ \frac{1}{2}\times \left(60-2L\right)=0 \end{gathered}$$

Solve the above equation for L:

$$\begin{gathered} \frac{1}{2}\times \left(60-2L\right)=0 \\ \left(60-2L\right)=0 \\ 2L=60 \\ =30 \end{gathered}$$

The obtained value of L is $30$.

Hence the length of the rectangular Patio to cover the maximum area is $30$ ft.

Thus, the corresponding width is:

$$\begin{gathered} W=\frac{\left(60-L\right)}{2} \\ =\frac{\left(60-30\right)}{2} \\ =\frac{30}{2} \\ =15 \end{gathered}$$

The value of W is $15$ ft

Hence, the dimensions of the ratio are in order to yield the maximum area is $30\text{ft},15\text{ft},15\text{ft}$.