Question

Determine the dimensions of a rectangular solid (with a square base) with maximum volume if its surface area is 529 square meters. Round each dimension to 4 decimal places

Answer 1

Answer:

$x=23/\sqrt{6}$,

$y= \frac{23}{6\sqrt{6}}$

Step-by-step explanation:

From the question we are told that

Surface area is 529 square meters

Generally the dimensions of the cuboid

$v=x^y$

$D=4xy+2x^2 =529$

$V=x^29(\frac{529-6x^2}{4x})$

$V' =529-6x^2/4$

$v'=0$

Therefore

$529=6x^2$

$x=\sqrt{\frac{529}{6} }$

$x=23/\sqrt{6}$

Mathematically

$v''=\frac{12x}{4}$

$v''=-3\frac{23}{\sqrt{6} }$

$y= \frac{529-2*(\frac{529}{6} }{4*\frac{23}{\sqrt{6} } }$

$y= \frac{23}{6\sqrt{6}}$

Therefore

$x=23/\sqrt{6}$

and

$y= \frac{23}{6\sqrt{6}}$