<em>blue</em>....... white....... <em>blue</em>....... <u>yellow</u>
red ........<em>green</em> .......red...........white
white........<u>yellow.</u>......<em>blue</em>.........<em>green</em>
<em /><em>blue</em><em> .........green ...... </em>red ........<u>.yellow</u>
Square means L=W
V=62.5=LWH
L=W so
V=62.5=HL^2
SA=2(L^2+2LH)
we have
V=62.5=HL^2
solve for H
divide both sides by L^2
62.5/L^2=H
sub that for H in other equation
SA=2(L^2+2L(62.5/L^2))
SA=2(L^2+125/L)
SA=2L^2+250/L
find minimum of 2L^2+250L^-1
take the derivitive
4L-250L^-2, or 2(2L^3-125)/L^2
find where it equals zero
it equals zero at L=2.5∛4
L=W
if we evaluate 2L^2+250/L at L=2.5∛4, the value is 75∛2
H=62.5/L^2
H=
![\frac{25 \sqrt[3]{2} }{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%20%5Csqrt%5B3%5D%7B2%7D%20%7D%7B4%7D%20)
dimentions are
L=2.5∛4
W=2.5∛4
H=
![\frac{25 \sqrt[3]{2} }{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%20%5Csqrt%5B3%5D%7B2%7D%20%7D%7B4%7D%20)
minimum surface area is 75∛2 in^2 or aprox 94.4941 in^2
Answer:
Step-by-step explanation:
![( \sqrt{5} )( \sqrt[3]{5}) \\ \\ = ({5}^{ \frac{1}{2} } ).({5}^{ \frac{1}{3} } ) \\ \\ = {5}^{ \frac{1}{2} + \frac{1}{3} } \\ \\ = {5}^{ \frac{3 + 2}{2 \times 3}} \\ \\ \huge \red{ \boxed{ = {5}^{ \frac{5}{6}} }}](https://tex.z-dn.net/?f=%28%20%5Csqrt%7B5%7D%20%29%28%20%5Csqrt%5B3%5D%7B5%7D%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%28%7B5%7D%5E%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%20%29.%28%7B5%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%7B5%7D%5E%7B%20%5Cfrac%7B1%7D%7B2%7D%20%20%2B%20%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%7B5%7D%5E%7B%20%5Cfrac%7B3%20%2B%202%7D%7B2%20%5Ctimes%203%7D%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Chuge%20%5Cred%7B%20%5Cboxed%7B%20%3D%20%7B5%7D%5E%7B%20%5Cfrac%7B5%7D%7B6%7D%7D%20%7D%7D)

I see I did something wrong in the 2nd part, just take the first question as the right one.