What are two positive numbers whose sum of the first number squared and the second number is 54 and the product is a maximum?

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Answer 1 · 2018-05-09T03:35:33

$3 \sqrt{2} and 36$ $3sqrt(2) and 36$

Let the numbers be $w$ $w$ and $x$ $x$ .

$x^{2} + w = 54$ $x^2 + w = 54$

We want to find

$P = w x$ $P = wx$

We can rearrange the original equation to be $w = 54 - x^{2}$ $w = 54 - x^2$ . Substituting we get

$P = (54 - x^{2}) x$ $P = (54 - x^2)x$

$P = 54 x - x^{3}$ $P = 54x - x^3$

Now take the derivative with respect to $x$ $x$ .

$P' = 54 - 3x^2$

Let $P' = 0$ .

$0 = 54 - 3x^2$

$3x^2 =54$

$x =+-sqrt(18) =+- 3sqrt(2)$

But since we're given that the numbers have to be positive, we can only accept $x =3sqrt(2)$ . Now we verify that this is indeed a maximum.

At $x = 3$ , the derivative is positive.

At $x = 5$ , the derivative is negative.

Therefore, $x =3sqrt(2)$ and $54 -(3sqrt(2))^2 = 36$ give a maximum product when multiplied.

Hopefully this helps!