Question

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

Answer 1

Solution:

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;

$\frac{dy}{dx}=0$

Then;

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

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;

$3\times3=9$

FINAL ANSWER: 9