Solve the problem.
-A private shipping company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 120 in. Suppose you want to mail a box with square sides so that its
Dimensions are h by h by w and it's girth is 2h + 2w. What dimensions will give the box its largest volume?
A) 20 in. $\times 20$ in. $\times 100$ in.
B) $\frac { 80 } { 3 }$ in. $\times \frac { 80 } { 3 }$ in. $\times 20$ in.
C) 40 in. $\times 20$ in. $\times 40$ in.
D) 20 in. $\times 20$ in. $\times 40$ in.

Question

Solve the problem.
-A private shipping company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 120 in. Suppose you want to mail a box with square sides so that its
Dimensions are h by h by w and it's girth is 2h + 2w. What dimensions will give the box its largest volume? 
A) 20 in. $\times 20$ in. $\times 100$ in.
B) $\frac { 80 } { 3 }$ in. $\times \frac { 80 } { 3 }$ in. $\times 20$ in.
C) 40 in. $\times 20$ in. $\times 40$ in.
D) 20 in. $\times 20$ in. $\times 40$ in.

Quizplus · Accepted Answer

Let h = w = x. Then the girth, 2h + 2w = 4x. Thus, we need 4x + 3x = 120, so x = 24. Therefore, the dimensions of the box are 24 in. $\times 24$ in. $\times 24$ in. which gives a volume of 13,824 cubic inches. To check that there are no larger dimensions, suppose the box has dimensions a by a by b, where a > b. Then it must be the case that $a^{2}+b^{2}+b^{2}+a^{2}\leq120$, or equivalently $2a^{2}+2b^{2}\leq120$. Dividing by 2, we obtain $a^{2}+b^{2}\leq60$, so $a^{2}\leq60$, or $a\leq\sqrt{60}$. Thus, the side of the square cannot be any larger than 24 inches, so this is the only choice that works.

Solve the Problem ×20\times 20×20 In ×100\times 100×100 In B) 803\frac { 80 } { 3 }380​

Solve the Problem $\times 20$ In $\times 100$ In
B) $\frac { 80 } { 3 }$