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March 2, 2015

March 2, 2015

Posted by **John** on Tuesday, November 27, 2012 at 6:01pm.

- Pre-calculus -
**Reiny**, Tuesday, November 27, 2012 at 6:13pmlet the side of the square to be cut out be x inches

resulting box is (12-2x) by (18-2x) by x

V = x(12-2x)(18-2x)= x(216 - 60x + 4x^2)

= 4x^3 - 60x^2 + 216x

dV/dx = 12x^2 - 120x + 216

= 0 for a max of V

x^2 - 10x + 18=0

solve for x

- Pre-calculus -
**Damon**, Tuesday, November 27, 2012 at 6:16pmlength = (18 - 2x)

width = (12 - 2x)

V = x (18-2x)(12-2x)

= x(4)(9-x)(6-x)

= 4 x (54-15x+x^2)

= 4 (54 x -15x^2 + x^3)

dV/dx = 0 for max

0 = 54 - 30 x + 3 x^2

0 = 18 - 10 x + x^2

x = [ 10 +/- sqrt (100-72) ]/2

[ 10 +/- 2 sqrt 7 ]/2

= 5 +/- sqrt 7

5+sqrt 7 is no good, more than half the width

so

5-sqrt 7 = 2.35

- Pre-calculus -
**Damon**, Tuesday, November 27, 2012 at 6:18pmLooks like calculus not pre-calculus to me by the way. I do not see how to do it without taking the derivative.

- Pre-calculus -
**John**, Tuesday, November 27, 2012 at 6:38pmwhat does sqrt mean?

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