And then I could draw-Īnd since it's open top, I can see through, I could see Open storage container, this open rectangular Find the cost of the materialįor the cheapest container. Material for the sidesĬosts $6 per square meter. Material for the baseĬosts $10 per square meter. The $4 per foot side Y=340/√(130)=29.820 feetĬontainer with an open top needs to have a volume Now to find C plug X into the cost equation. In either case we don't have to do that in this problem, so moving on. You could either do the second derivative test to find out whether they are maximums or minimums, or you could simply plug into the cost equation and find out which one is cheaper. If there were multiple positive X's it would be simple to test. Right away you can discard the negative answer because X is a length and you cannot have negative length, so X=2√(130). Next, set dc/dx equal to 0 this will allow you to find the x value for all mins and maxes in the original equation. Simply isolate the Y variable on one side of the first equation so you can substitute the expression it's equal to into the second equation.Ĭhange the 6240/x to 6240X^-1 to make deriving easier.

To move on you need a single equation with only 2 variables. You can then assign the $6 cost to one variable and the $4 cost to the other.

Perimeter is equal to twice the length + twice the width. the cost of the materials is going to be determined by the perimeter of the space. You also know that c is the combined cost of the materials. X=width of the space, Y=length of the space, and C=cost of materials.īecause you know that the area is 780 square feet, you know that 780 is the product of x and y.

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