a. The volume of a rectangular package is given by V = LWH, where L is the length, W is the width, and H is the height. From the problem statement, we have:
L = W + 10
H = W - 12
Substituting these expressions into the formula for volume, we get:
V = (W + 10)(W)(W - 12)
V = W^3 - 2W^2 - 120W
So the polynomial function that calculates the volume of a package with width, w, is:
V(w) = w^3 - 2w^2 - 120w
b. To find the dimensions that will give us the maximum allowable volume, we need to find the maximum value of the volume function V(w), subject to the constraint that the volume cannot exceed 4800 cubic inches. We can use the method of calculus to find the maximum value.
First, we need to find the derivative of V(w):
V'(w) = 3w^2 - 4w - 120
Setting V'(w) equal to zero and solving for w, we get:
3w^2 - 4w - 120 = 0
(w - 6)(3w + 20) = 0
The only positive solution is w = 6 inches (since the width cannot be negative). To verify that this is a maximum, we need to check the sign of the second derivative of V(w):
V''(w) = 6w - 4
When w = 6, V''(w) = 32, which is positive. Therefore, the function V(w) has a local minimum at w = 6, which is also the global maximum (since V(w) approaches infinity as w goes to infinity).
So the dimensions that will give us the maximum allowable volume of 4800 cubic inches are:
Width = 6 inches
Length = 16 inches (since L = W + 10 = 16)
Height = -6 inches (since H = W - 12 = -6, we interpret this as a package with the top and bottom sides flipped, so that the height is positive)
Therefore, the maximum volume of the package that meets the size requirement is:
V = LWH = (16)(6)(6) = 576 cubic inches.
Hope this helped!