SOLUTION: A box is to be constructed from a sheet of cardboard that is 20 cm by 60 cm, by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum

Algebra ->  Finance -> SOLUTION: A box is to be constructed from a sheet of cardboard that is 20 cm by 60 cm, by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum       Log On


   

Question 987179: A box is to be constructed from a sheet of cardboard that is 20 cm by 60 cm, by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum volume this box could have?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The 20cm side would become 20-2x long.
The 60cm side would become 60-2x long.
The total volume of the box would be V=%2820-2x%29%2860-2x%29x
V=4x%5E3-160x%5E2%2B1200x
To find the maximum volume, take the derivative and set it equal to zero.
dV%2Fdx=12x%5E2-320x%2B1200
3x%5E2-80x%2B300=0
3%28x%5E2-%2880%2F3%29x%29%2B300=0
3%28x%5E2-%2880%2F3%29x%2B%2840%2F3%29%5E2%29%2B300=3%2840%2F3%29%5E2
3%28x-40%2F3%29%5E2=3%281600%2F9%29-300
3%28x-40%2F3%29%5E2=1600%2F3-900%2F3
3%28x-40%2F3%29%5E2=700%2F3
%28x-40%2F3%29%5E2=700%2F9
x-40%2F3=0+%2B-+%2810%2F3%29sqrt%287%29
x=40%2F3+%2B-+%2810%2F3%29sqrt%287%29
Of the two values, only x=40%2F3-%2810%2F3%29sqrt%287%29 is in the range of x which can only have values between 0 and 10 since 20-2x must remain positive.
So then the maximum volume that the box can have is,
V%5Bmax%5D=%2856000sqrt%287%29-80000%29%2F27