Question

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. 

Answer 1

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)

$S=x^2+y^3= \Bigg( \cfrac{4\sqrt{3}}{y} \Bigg)^2+y^3= \cfrac{48}{y^2}+y^3 \\ \cfrac{dS}{dy} = -\cfrac{96}{y^3} +3y^2$

To find the minimum we have to make $\cfrac{dS}{dy} =0$ and then solve for y:

$-\cfrac{96}{y^3} +3y^2=0 \\ \implies 3y^2=\cfrac{96}{y^3} \, \therefore y^5=\cfrac{96}{3}=32=2^5 \\ \implies y=2 \\ \therefore x= \cfrac{4\sqrt{3}}{2}=2\sqrt{3}$

In conclusion, the numbers $x=2\sqrt{3}$ and $y=2$ satisfy the relation given ($xy=4\sqrt{3}$ ) and the value for S is minimum.

Reference:

• Thomas, G. B., & Finney, R. L. (1961). Calculus. Addison-Wesley Publishing Company.