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Study Guides > Business Calculus

Putting It Together

Summary

Back to our box problem at the beginning. If we let x = side of a square that is to be cut out, then the volume of the box we will wind up with is given by

V = 144x − 24x2 + x3

Since we are wanting to find the maximum volume, we know to take the derivative of this to calculate our minimums and maximums. By performing our first derivative and setting it equal to zero:

V1 = 144 − 48x + 3x2

We find we have critical points at x = 4 and x = 12 Using our second derivative test, we discover that 4" squares will give us the maximum volume of the box that we are seeking.

Licenses & Attributions

CC licensed content, Original

  • Provided by: Columbia Basin College Authored by: Paul Jones. License: CC BY: Attribution.