'Simple' Optimization Problem

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SUMMARY

The optimization problem involves determining the height of an open box formed by cutting corners from a 14-inch by 10-inch piece of cardboard to maximize volume. The volume function is defined as V = (10-2x)(14-2x)x, leading to the derivative dv/dx = 12x^2 - 96x + 140. The critical points calculated are x = 6.08 and x = 1.92, with only x = 1.92 being feasible for the box's height. The discussion highlights the importance of verifying calculations and understanding the implications of critical points in optimization problems.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly derivatives
  • Familiarity with volume formulas for rectangular prisms
  • Knowledge of solving quadratic equations
  • Ability to analyze critical points for optimization
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  • Study the application of the First Derivative Test in optimization problems
  • Learn about the method of Lagrange multipliers for constrained optimization
  • Explore volume optimization problems involving different geometric shapes
  • Practice solving real-world optimization scenarios using calculus
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Students preparing for calculus exams, educators teaching optimization techniques, and anyone interested in applying calculus to practical problems in geometry and volume maximization.

ObviousManiac
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Just reviewing for a chapter test... I've always found optimization problems easy, but I don't have answers for these review questions so I thought I'd check my work on here.

Homework Statement


A piece of cardboard is 14 inches by 10 inches and you are going to cut out the corners and fold up the sides to form an open box. Determine the height of the box that will give a maximum volume.

Homework Equations



V = lwh

The Attempt at a Solution



Using
(10-2x) as l
(14-2x) as w
and x as the height:

V = (10-2x)(14-2x)x = 4x^3 - 48^2 +140x

dv/dx = 12x^2 - 96x + 140

0 = 12x^2 - 96x +140

x = 6.08, 1.92 or 1/3(12±√39)

... this is where I get lost. I feel like I messed it up somewhere because I get weird numbers for x. Technically, the 6.08 number is impossible, so x would have to equal 1.92. But still, I feel uncomfortable with these numbers.
 
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What makes you uncomfortable about the answer you got?

Would it make you more comfortable to compare the volume if the height were slightly above or below the height you calculated to the volume you got?
 
jamesrc said:
Would it make you more comfortable to compare the volume if the height were slightly above or below the height you calculated to the volume you got?

haha, yes, actually. It helped quite a bit. I think I'm just too tired from too many late nights of studying... thanks for the help.
 

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