How do you find the dimensions and maximum area of a rectangle whose perimeter is 24 inches?

Question

Nathan Axel · Accepted Answer

A #6xx6# square with area 36 square inches

The rectangle has sides h and l, perimeter p and area a: (a) #p=2h+2l=24# (b) #a=h*l# (a) #h=12-l# (b) #a=(12-l)*l= 12l-l^2# The maximum of #a# occurs when the first derivative is zero: #( 12l-l^2)'= 12-2l=0# #l=12/2=6# #h=12-6=6#

Abigail Allen · Answer

undefined To find the dimensions and maximum area of a rectangle with a given perimeter of 24 inches, we can use the formula for the perimeter of a rectangle, which is $ P = 2l + 2w $, where $ l $ is the length and $ w $ is the width.

Given $ P = 24 $, we can rearrange the formula to solve for one of the variables. Let's solve for $ l $:

$$ 24 = 2l + 2w $$

$$ 12 = l + w $$

$$ l = 12 - w $$

Now, we can express the area $ A $ of the rectangle in terms of $ w $. The area of a rectangle is given by $ A = lw $. Substitute the expression for $ l $ from above into the area formula:

$$ A = (12 - w)w $$

$$ A = 12w - w^2 $$

To find the maximum area, we can take the derivative of the area formula with respect to $ w $, set it equal to zero, and solve for $ w $. Then, we can use this value of $ w $ to find the corresponding value of $ l $. Finally, we can calculate the maximum area using the dimensions $ l $ and $ w $.

Taking the derivative of $ A $ with respect to $ w $:

$$ \frac{dA}{dw} = 12 - 2w $$

Setting the derivative equal to zero:

$$ 12 - 2w = 0 $$

$$ 2w = 12 $$

$$ w = 6 $$

Now that we have found $ w = 6 $, we can find $ l $:

$$ l = 12 - w $$

$$ l = 12 - 6 $$

$$ l = 6 $$

So, the dimensions of the rectangle are $ l = 6 $ inches and $ w = 6 $ inches.

The maximum area $ A $ is:

$$ A = lw $$

$$ A = 6 \times 6 $$

$$ A = 36 \text{ square inches} $$

Therefore, the maximum area of the rectangle is $ 36 $ square inches.