SOLUTION: A closed box, whose length is thrice its width, is to have a surface of 200 square cm. Find the maximum Volume of the box?

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Question 1172105: A closed box, whose length is thrice its width, is to have a surface of 200 square cm. Find the maximum Volume of the box?
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
Total Surface Area = 6x^2 (top and bottom) + 8xh (4 sides) = 200 cm^2
Tot SA = 6x^2 + 8xh = 200
Finding 'x' value of 'h'
8xh = 200 - 6x^2
h = (200 - 6x^2)/8x
Cancelling 2 top and bottom
h = (100 - 3x^2)/4x
Volume = length * breadth * height (Using * for multiply)
V= 3x * x * h
V = 3x^2h
Bringing in 'x' value for 'h'
V(x) = 3x^2(100 - 3x^2)/4x
3/4x (100 - 3x^2)
V(x) = (300/4)x - 9x^3/4
Cancel down
V(x) = 75x - 9x^3/4
Differentiate
V'(x) = 75 - 27x^2/4
Make V'(x) = 0
75 - 27x^2/4 = 0
-27x^2/4 = - 75
Multiply both sides by -1
27x^2/4 = 75
27x^2 = 300
x^2 = 300/27 or 100/9
x = 10/3.
......................................................... - 10/3 +
Nature Table.75 - 27x^2/4 ...... + 0 -
........................................................... / - \
Maximum
height = (100 - 3x^2)/4x
Height = [100 - 3(10/3)^2] / 4 (10/3)
Height = 5 cm
Total Volume = length * breadth * height
Total volume = 3(10/3) * (10/3) * 5
Total Volume = 10 * 10/3 * 5
Total volume = 166.66 cm^3
Hope this helps :-)