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Optimization calculus khan academy
Optimization calculus khan academy












optimization calculus khan academy

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.

optimization calculus khan academy

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.














Optimization calculus khan academy