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how do i begin
Do you know if all 1200 ft^2 of material is to be used? or is it one of those"cut out the corners" kind of problems?
the container, if all material is to be used would have a surface area of: 4(side area) + (base area) = 1200
you just need the sum of the areas of the surfaces to add up to 1200sqft. so (area of base)+4×(area of one side) = 1200 sqft
4xy + x^2 = 1200 , does that sound right?
Volume = x^2 y
solve for one variable in terms of the other, plug it into the volume equation, and derive :)
y = (1200 -x^2)/4x
and dont let the "hero" title fool ya.... I am after all an idiot in disguise :)
V = (x^2/4x)(1200-x^2)
V = (1/4) (1200-x^3) if I did it right...
*1200x - x^3
dV = (1/4)(1200 - 3x^2) maybe make that equal to 0 to get the critical numbers
Oh, I misread the problem.. yes, put y in terms of x, y = (1200 -x^2)/4x, then substitute into volume V = x^2 y = x^2 (1200 -x^2)/4x = x(1200-x^2)/4 = x(300-(x^2)/4), then find the maximum volume
1200 = 3x^2
x = sqrt400)
If you've taken calculus you can do that my finding the zeros of the derivative, otherwise I think you'll have to test sample values and just find the x that yields the greatest V
x=20 i think plug that back in to your "y" equation to ge the value for y
y = (1200 -20^2)/4(20) y = (1200 - 400)/ 80 y = 800/80 = 80/8 = 10/1 y = 10 and x = 20 if i did it all correctly
determine if 20x20x10 is greater than 10x10x20 and youve got your answer
or.... ignore that last comment...its probably my stupidity talking :)
20x20 for the base, and 10 high fits all the requirements
but, do you see how we got it?
square roots :)