Question

A postal service will accept packages only if the length plus girth is no more than 264 inches. (See the figure.) Assuming that the front face of the package (as shown in the figure) is square, what is the largest volume package that the postal service will accept?

Answer 1

so.. .we know the front face of the package is a square, that means the width and height are the same value, notice the picture below

thus... .if the length = l = 264 - 4x

well, the volume is length*width*height

thus
`\bf \textit{volume of a rectangular prism}\\\\ V=length\cdot width\cdot height\qquad \begin{cases} width=x\\ height=x\\ length=l\\ ------\\ l=264-4x \end{cases}\implies V=l\cdot x\cdot x \\\\\\ V=l\cdot x^2\implies V(x)=(264-4x)x^2\implies V(x)=264x^2-4x^3`

so hmm, take the derivative of V(x), zero it out, check the critical points for any minima in the interval of (0, 264), since neither the height or width can be 0 or 264, but in between

Answer 2

V_max = 170,368 in^3

Explanation:

Given:-

- The length of the package = L

- The side-length of the square base = x

- The girth of the package = 4x

- L + 4x ≤ 264

Find:-

what is the largest volume package that the postal service will accept?

Solution:-

- The volume of the postal service package is given by the volume of a cuboid such that, volume of the package V is:

V ( x , L) = x^2 * L

- We see that volume is a function of both side length of square base (x) and the length of the package (L). We will use the constraint given and determine that the relationship between L and x. We have:

L + 4x ≤ 264

L ≤ 264 - 4x

- We will use the developed relationship and substitute into the volume expression.

V(x) = x^2* ( 264 - 4x )

V(x) = -4x^3 + 264 x^2

- To find the largest volume we will have to maximize our function of Volume V (x). Taking first derivative of the V (x) wrt to side length of the square base (x).

V'(x) = -12x^2 + 528x

- Set the first derivative V' (x) equal to zero and solve for x:

0 = x ( -12x + 528 )

0 = x , -12x + 528 = 0

x = 44 in

- The critical value for which the largest or the smallest volume exist are x = 0 and x = 44 in. From basic intuition we can see that x = 0 is the minimizing critical value while x = 44 in is the maximizing one. Compute the largest volume by plugging the maximum critical value:

V(x) = -4x^3 + 264 x^2

V ( 44 ) = -4 (44)^3 + 264 (44)^2

V ( 44 ) = -340,736 + 511,104

V_max = 170,368 in^3