Asked by John
You are given a 12"x18" piece of construction paper. You are to cut a square out of each corner in order to create a 3-dimensional open-top box. What size squares would you need to cut in order to maximize the volume of the box?
Answers
Answered by
Reiny
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
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
Answered by
Damon
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
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
Answered by
Damon
Looks like calculus not pre-calculus to me by the way. I do not see how to do it without taking the derivative.
Answered by
John
what does sqrt mean?
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