# Minimum cost word problem (calculus: optimization)

• Feb 24th 2014, 08:51 PM
angie427
Minimum cost word problem (calculus: optimization)
An open rectangular box (no top) with a volume 9 cubic meters has a square base. The base has side x m. The equation of surface area of the box is x2 + 36/x.
The base costs 1 cents/square m and the sides cost 1.6 cents/square m. I found the expression for the cost as a function of x to be x2 (1) + 57.6/x.

I can't figure out how to calculate the value of x to achieve minimum cost??? It has to be calculated with the second derivative test.
• Feb 25th 2014, 08:40 AM
hollywood
Re: Minimum cost word problem (calculus: optimization)
Your formula for cost looks correct. You need to take the derivative and set it equal to zero to find the minimum.

The second derivative test is used to determine if your answer is a minimum or maximum. In this case, it is obviously a minimum, so no second derivative test is required. Unless it's a second-derivative test exercise, of course.

- Hollywood
• Feb 25th 2014, 08:29 PM
angie427
Re: Minimum cost word problem (calculus: optimization)