A carpenter is building a rectangular room with a fixed perimeter of 92 ft. What dimensions would yield the maximun perimeter

Question

 Need some help with my algebra homework. Word problems are difficult for me.

Answer 1

Hi Tracy, if the room has a fixed perimeter of 92 feet, that would yield the maximum perimeter. What we can find out is what the maximum length and width of the room would be that would give the maximum area and have a perimeter of 92 feet.

 

We can use the equation for the perimeter of a rectangle: P = 2L + 2W.

 

Using the formula:

 

We could have a room 1 foot wide by 45 feet long with an area of 45 square feet. Not much of a room.

 

We could have a room 2 feet wide by 44 feet long with an area of 88 square feet. Again, not much of a room.

 

We could continue on until we found the largest room that fit the given perimeter.

 

Or we could find the room with the largest area using algebra.

 

Start by substituting for P:

 

P = 92 feet

 

2L + 2W = 92

 

Divide both sides by 2:

 

L + W = 46

 

Solve for L:

 

L = -W + 46

 

The formula for the area of a rectangle is:

 

A = L * W

 

Substitute for L:

 

A = (-W + 46) * W

 

A = -W² + 46 W

 

Now we have a quadratic. We can use the information from the vertex to answer our question:

 

Since the W² term is negative, we have a maximum.

 

The vertex of a parabola is: (h, k), where h is: -b / 2a.

 

b = 46

 

a = -1

 

h = -46 / (2 * -1)

 

h = -46 / -2, or h = 23

 

k = -W² = 46W

 

Substitute h for W gives:

 

k = -(23²) + 2 * 23

 

k = -529 + 1058

 

k = 529

 

The maximum area of the rectangle is 529 square feet.

 

We know the width, W, equals 23 feet. To find the length, we need to solve: A = L * W for L after substituting for A and W.

 

529 = L * 23

 

Divide both sides by 23:

 

L = 23

 

The length of the rectangle is 23 feet.

 

What we found is the maximum length and width of a rectangle if we have a perimeter of 92 feet.

 

The maximum rectangle is actually a square with each side 23 feet in length.

 

Questions?