Lesson Maximize the volume of an open box

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Maximize the volume of an open box

Problem 1

A tinsmith wants to make a small planter from a 38cm x 22 cm sheet of copper.
He’ll form it by cutting equally sized squares from each of the four corners of the sheet,
folding up the resulting flaps to form the sides of the planter, and then shouldering the four vertical edges.
What is the maximum possible volume of the planter?

Solution

After folding up the resulting flaps, the base of the planter has dimensions
(38-2x) cm by (22-2x) cm and the height is x cm.


Thus the volume is  

    V(x) = (38-2x)*(22-2x)*x  cm^3.


To find the maximum V(x), take the derivative and equate it to zero

    0 = V'(x) = -2*(22-2x)*x + (38-2x)*(-2)*x + (38-2x)*(22-2x) = 

          = -44x + 4x^2  - 76x + 4x^2     + 38*22 - 44x - 76x + 4x^2 = 

          = 12x^2 - 240x + 836.


It gives  

    3x^2 - 60x + 209 = 0.


The roots are (use the quadratic formula)

     = ,

or

     = 15.5076,   = 4.49243.


The larger root is TOO big and, therefore, is not the solution to the problem.

The other value, 4.49243, is the approximate solution for x.


Thus the maximum volume is  

     = 4.49243*(38-2*4.49243)*(22-2*4.49243) = 1696.504 cm^3 = 1697 cm^3,
                                       rounded to the closest integer value, in cm^3.    ANSWER

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Use this file/link ALGEBRA-II - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-II.

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