The volume of an open-topped box with a square base is 245 cubic centimeters. find the height of the box that uses the least material. can someone verify for me that the answer is 3.942 centimeters for the height?

X = side length of square base
h = height

area of base = x^2
volume = height*(area of base)
volume = h*x^2
245 = h*x^2
h = 245/(x^2)



Surface area

SA = x^2+4*h*x
SA = x^2+4*(245/(x^2))*x
SA = x^2+4*(245/x)
SA = x^2+4*(245/x)
S(x) = x^2+(980/x)

Goal is to minimize S(x)



S(x) = x^2+980/x
S(x) = x^2+980x^(-1)
S'(x) = 2x-980x^(-2)
S'(x) = 2x-980/(x^2)
S'(x) = (2x^3)/(x^2)-980/(x^2)
S'(x) = (2x^3-980)/(x^2)
0 = (2x^3-980)/(x^2)
2x^3-980 = 0
2x^3 = 980
x^3 = 980/2
x^3 = 490
x = 490^(1/3)
x = 7.883735

The critical value is roughly x = 7.883735

Use the first derivative test
 Plug in x = 7
S'(x) = (2x^3-980)/(x^2)
S'(7) = (2(7)^3-980)/(7^2)
S'(7) = -6
then plug in x = 8
S'(x) = (2x^3-980)/(x^2)
S'(8) = (2*8^3-980)/(8^2)
S'(8) = 0.6875
The derivative changes sign as you move through the critical value. Since we change from negative to positive, this shows we have a local min at roughly x = 7.883735

Plug this x value into the h(x) function to get the height
h(x) = 245/(x^2)
h(7.883735) = 245/((7.883735)^2)
h(7.883735) = 3.941867
which rounds to 3.942

So you have the correct answer. Nice work getting two in a row.