Posted by John on Tuesday, November 27, 2012 at 6:01pm.
let 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
length = (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
Looks like calculus not pre-calculus to me by the way. I do not see how to do it without taking the derivative.
what does sqrt mean?
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