Question

Kern Shipping Inc. has a requirement that all packages must be such that the combined length plus the girth (the perimeter of the cross section) cannot exceed 99 inches. Your goal is to find the package of maximum volume that can be sent by Kern Shipping. Assume that the base is a square.
a. Write the restriction and objective formulas in terms of x and y. Clearly label each.
b. Use the two formulas from part (a) to write volume as a function of x, V(x). Show all steps.

Answer 1

Answer:

Step-by-step explanation:

From the given information:

a)

Assuming the shape of the base is square,

suppose the base of each side = x

Then the perimeter of the base of the square = 4x

Suppose the length of the package from the base = y; &

the height is also = x

Now, the restriction formula can be computed as:

y + 4x ≤ 99

The objective function:

i.e maximize volume V = l × b × h

V = (y)*(x)*(x)

V = x²y

b) To write the volume as a function of x, V(x) by equating the derived formulas in (a):

y + 4x ≤ 99   --- (1)

V = x²y          --- (2)

From equation (1),

y ≤ 99 - 4x

replace the value of y into (2)

V ≤ x² (99-4x)

V ≤ 99x² - 4x³

Maximum value V = 99x² - 4x³

At maxima or minima, the differential of $\dfrac{d }{dx}(V)=0$

$\dfrac{d}{dx}(99x^2-4x^3) =0$

⇒ 198x - 12x² = 0

$12x \Big({\dfrac{33}{2}-x}}\Big)=0$

By solving for x:

x = 0 or x = $\dfrac{33}{2}$

Again:

V = 99x² - 4x³

$\dfrac{dV}{dx}= 198x -12x^2 \\ \\ \dfrac{d^2V}{dx^2}=198 -24x$

At x = $\dfrac{33}{2}$

$\dfrac{d^2V}{dx^2}\Big|_{x= \frac{33}{2}}=198 -24(\dfrac{33}{2})$

$\implies 198 - 12 \times 33$

= -198

Thus, at maximum value;

$\dfrac{d^2V}{dx^2}\le 0$

Recall y = 99 - 4x

when at maximum x = $\dfrac{33}{2}$

$y = 99 - 4(\dfrac{33}{2})$

y = 33

Finally; the volume V = x² y is;

$V = (\dfrac{33}{2})^2 \times 33$

$V =272.25 \times 33$

V = 8984.25 inches³