1. ## Help

[FONT='Cambria','serif']There are 2 positive numbers who have a sum of 9. Determine the maximum value of the product of one number and the square of the other number.[/FONT]

2. Originally Posted by someone21
[FONT='Cambria','serif']There are 2 positive numbers who have a sum of 9. Determine the maximum value of the product of one number and the square of the other number.[/FONT]
Let x, y denote the two positive numbers.

Then you know:

$\displaystyle x+y=9~\implies~x = 9-y$ ...... [1]

$\displaystyle p = x \cdot y^2$ ...... [2]

Plug in the term of [1] into [2]:

$\displaystyle p(y) = (9-y) \cdot y^2 = -y^3+9y^2$

Calculate the first derivative and solve the equation $\displaystyle p'(y) = 0$ for y.

I've got y = 0 or y = 6 and therefore the maximum value of the product is 108.

3. Originally Posted by someone21
[FONT='Cambria','serif']There are 2 positive numbers who have a sum of 9. Determine the maximum value of the product of one number and the square of the other number.[/FONT]
$\displaystyle a+b = 9$. This means $\displaystyle ab^2 = (9 - b)b^2 = 9b^2 - b^3 = f(b)$

Maxima of f(b) is when f'(b) = 0 . So $\displaystyle f'(b) = 18b - 3b^2 = 0 \Rightarrow b = 6$. $\displaystyle b = 0$ is the minima.
Thus $\displaystyle a = 3$ and the maximum product is $\displaystyle 3.6^2 = 108$

4. Originally Posted by someone21
[FONT='Cambria','serif']There are 2 positive numbers who have a sum of 9. Determine the maximum value of the product of one number and the square of the other number.[/font]
let $\displaystyle x$ be one of the numbers. so the other can be represented as $\displaystyle 9 - x$

Let $\displaystyle P(x)$ be the function representing the product as stated WLOG: $\displaystyle P(x) = x^2(9-x) = 9x^2 - x^3$

so, $\displaystyle P'(x) = 18x - 3x^2 = 3x(6-x)$ and we want the maximum, thus we need $\displaystyle x$ such that $\displaystyle P'(x) = 0$

therefore, either $\displaystyle x=0$ or $\displaystyle x=6$ but note that x is a positive number, therefore we take $\displaystyle x=6$.

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optional: if $\displaystyle P''(x_0) < 0$, then we are sure that $\displaystyle x_0$ gives the maximum value for $\displaystyle P(x)$

$\displaystyle \Rightarrow P''(x) = 18 - 6x$ and at $\displaystyle x=6$, we have $\displaystyle P''(6) = 18 - 36 = -18 < 0$.