A rectangle has a perimeter of 100 inches, and one side has length x. Express the area of the

rectangle as a function of x.

Use the function in part a to find the dimensions of the rectangle with permitter 100 inches and the largest possible area

a. A = 50x - x² b. length = 25 inches and width = 25 inches and the maximum area is 625 in²

Step-by-step explanation:

a. The perimeter of a rectangle P = 2(L + W) where L = length and W = width. Now, given that P = 100 inches and W = x, substituting these into the equation, we have

P = 2(L + W)

100 = 2(L + x)

dividing both sides by 2, we have

100/2 = L + x

50 = L + x

making L subject of the formula, we have

L = 50 - x

Now, the are of a rectangle A = LW. Substituting the values of L and W, we have

A = LW

A = (50 - x)x

A = 50x - x²

b. To find the largest possible area of rectangle with perimeter 100 inches, we differentiate A and equate it to zero to find the value of x that maximizes A.

So, dA/dx = d(50x - x²)/dx

dA/dx = d50x/dx - dx²/dx

dA/dx = 50 - 2x

dA/dx = 0 ⇒ 50 - 2x = 0

50 = 2x

dividing both sides by 2, we have

x = 50/2

x = 25

To find it this gives maximum value for A, we differentiate A twice.

d²A/dx² = d(50 - 2x)/dx

d²A/dx² = d50/dx - d2x/dx

d²A/dx² = -2

Since d²A/dx² = -2 < 0, so x = 25 gives maximum value for the area, A.

Since W = x = 25 in and L = 50 - x. So, L = 50 - 25 = 25 in

So, the maximum area A = LW = Lx = 25 in × 25 in = 625 in²

The dimension with perimeter 100 inches that give maximum area are length = 25 inches and width = 25 inches and the maximum area is 625 in²

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