The product of two positive numbers is 4 √ 3. Find the numbers so that the sum S of the square of one and the cube of the other is as small as possible.
First, we have the following relation that will help us to solve the equation that has to be minimized:
"xy=4\\sqrt{3} \\implies x=\\cfrac{4\\sqrt{3}}{y}; x>0, y>0"
Then, we use the formula for the sum and we substitute x to find an equation in terms of the other variable (y)
"S=x^2+y^3= \\Bigg( \\cfrac{4\\sqrt{3}}{y} \\Bigg)^2+y^3= \\cfrac{48}{y^2}+y^3\n\\\\ \\cfrac{dS}{dy} = -\\cfrac{96}{y^3} +3y^2"
To find the minimum we have to make "\\cfrac{dS}{dy} =0" and then solve for y:
"-\\cfrac{96}{y^3} +3y^2=0\n\\\\ \\implies 3y^2=\\cfrac{96}{y^3} \\, \\therefore y^5=\\cfrac{96}{3}=32=2^5\n\\\\ \\implies y=2\n\\\\ \\therefore x= \\cfrac{4\\sqrt{3}}{2}=2\\sqrt{3}"
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