Find the dimensions of a rectangular box of maximum volume such that the sum of the lengths of its 12 edges is
A) , ,
B) 4, 8, 16
C) 32, , 16
D) , ,
E) 32, 32, 32

Question

Find the dimensions of a rectangular box of maximum volume such that the sum of the lengths of its 12 edges is 
A) , , 
B) 4, 8, 16 
C) 32, , 16 
D) , , 
E) 32, 32, 32

Quizplus · Accepted Answer

Let the dimensions of the box be $l$, $w$, and $h$. Then we have the equations $2(l+w+h)=11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11$ and $V=lwh$. We can solve for $h$ in the first equation to get $h=\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{2}-l-w$. Substituting this into the equation for the volume gives $V=lw\left(\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{2}-l-wight)$. Expanding and completing the square gives $V=-\left(l-\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{4}ight)^2-\left(w-\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{4}ight)^2+\frac{11ea8a3b_b734_de05_9fff_676b7f4c4112_TB6897_11}{16}$. Thus, the maximum value of $V$ occurs when $l=w=\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{4}$, which gives $h=\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{2}-2\cdot\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{4}=\frac{11ea8a3b_b734_b6f0_9fff_45587bee33ed_TB6897_11}{4}$. Thus, the dimensions of the box are $\boxed{	extbf{(D) }11ea8a3b_b734_de05_9fff_676b7f4c4112_TB6897_11 , 11ea8a3b_b734_de06_9fff_4f52df677ecc_TB6897_11 , 11ea8a3b_b734_de07_9fff_bba7332a0a3d_TB6897_11}$.

Find the Dimensions of a Rectangular Box of Maximum Volume