Question

Use calculus to find the two non-negative real numbers with a sum of 6 have the smallest possible product.

Answer 1

Let x and y be the non-negative real numbers. Then;

`\begin{gathered} x+y=6 \\ y=6-x\ldots\ldots\ldots\text{.equation}1 \end{gathered}`

Also, then the product of the two non-negative real numbers is;

`\begin{gathered} xy=x(6-x) \\ xy=6x-x^2 \end{gathered}`

At minimum;

`(dy)/(dx)=0`

Then;

`\begin{gathered} (d(xy))/(dx)=6-2x=0 \\ 6=2x \\ \text{Divide both sides by 2;} \\ (6)/(2)=(2x)/(2) \\ x=3 \end{gathered}`

Put the value of x in equation 1, we have;

`\begin{gathered} y=6-x \\ y=6-3 \\ y=3 \end{gathered}`

Then, the smallest possible product of the two non-negative real numbers is;

`3*3=9`

FINAL ANSWER: 9