Answer to Question #262174 in Calculus for Amit

Question

Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.

Answer 1

Let $x{\scriptscriptstyle 1} = a, x{\scriptscriptstyle 2}=b$

We have to find such values a and b(a + b = 9) that $ab^2\to max$

Since a + b = 9, then a = 9 - b, hence $ab^2=(9-b)b^2=9b^2-b^3$ (0 < b < 9). According to the condition, b = 0 and b = 9 also should be included, but they are obviously doesn't maximize the given function cause it will be equal to 0

Now we have to find the maximum of the function on the interval. Let's find its derivative:

$(9b^2-b^3)'=18b-3b^2$

Find the critical points

$18b-3b^2=0\implies 3b=18\implies b=6$. To the left of b = 6 the derivative is positive, to the right it is negative, that means b=6 is the point of maximum and $ab^2=(9-b)b^2=96^2-6^3=108$

Those numbers are 3 and 6, the maximum value is 108.