Asked by Nisa
A square-bottomed box with no top has a fixed volume, V. What dimensions minimize the surface area?
Answers
Answered by
Reiny
let each side of the base be x
let the height be y
volume is V, a constant
x^y = V
y = V/x^2
SA = x^2 + 4xy
= x^2 + 4xV/c^2 = x^2 + 4V/x
d(SA)/dx = 2x - 4V/x^2 = 0 for min of SA
4V/x^2 = 2x
x^3 = 2V
x = (2V)^(1/3)
take it from there
let the height be y
volume is V, a constant
x^y = V
y = V/x^2
SA = x^2 + 4xy
= x^2 + 4xV/c^2 = x^2 + 4V/x
d(SA)/dx = 2x - 4V/x^2 = 0 for min of SA
4V/x^2 = 2x
x^3 = 2V
x = (2V)^(1/3)
take it from there
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.