Answer:
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
