Calculus min max problem - need help with the derivative.?

Question

One number squared times another number cubed is a maximum. Their sum is 20. How do I find?
They are both positive numbers. I understand it is a max question, and to use derivatives, how do I set up & solve?

By trial & error, I believe the answer is 9 & 11. (9^2 * 11^3). I keep getting 12 as one number.

Any help is appreciated, thanks.

Answer 1

a+b=20
so a=a
and b=20-a
so (20-a)^2*a^3=y and you want to maximize y
using the product rule dy/da= 3a^2*(20-a)^2 -2(20-a)*a^3=0
This simplifies to 3(20-a)-2a=0 (remember that a can not equal 0 or 20) so 60-5a=0 or 60=5a so a=12 and b=8.

Answer 2

Let x and y be both of the numbers. It is given that their sum is equal to 20, so

x + y = 20

We want to maximize (x^2)(y^3), so let's define

M = (x^2)(y^3)

First, express y in terms of x using the first equation. x + y = 20, so y = 20 - x. Substituting this, we get

M = (x^2)(20 - x)^3

Now we can define this to be our maximize function, M(x).

M(x) = (x^2)(20 - x)^3

Take the derivative. Using the product rule,

M'(x) = (2x)(20 - x)^3 + (x^2)(3)(20 - x)^2 (-1)
M'(x) = (2x)(20 - x)^3 - (3x^2)(20 - x)^2

Rather than expand it all out, let's be clever about this and just factor out the greatest common monomial out of these complex product of terms.

M'(x) = x(20 - x)^2 [ 2(20 - x) - 3x ]
M'(x) = x(20 - x)^2 [ 40 - 2x - 3x ]
M'(x) = x(20 - x)^2 [ 40 - 5x ]
M'(x) = x(20 - x)^2 (5)(8 - x)
M'(x) = 5x(20 - x)^2 (8 - x)

Now make M'(x) = 0 to solve for the critical numbers.

0 = 5x(20 - x)^2 (8 - x)

Making each factor equal to 0,

5x = 0
(20 - x)^2 = 0
8 - x = 0

Leading to the solutions x = {0, 20, 8}

Now we determine where we have a local maximum. Make a number line consisting of the critical numbers.

. . . . . . . . . (0) . . . . . . . . . . . (8) . . . . . . . . . . . (20) . . . . . . . . . .

Here is where we test each region to determine the sign of M'(x). All we have to do is test one value in each region.

Test x = -1:
M'(x) = 5x(20 - x)^2 (8 - x)
M'(-1) = [negative] [positive] [positive] = negative.
Mark the region as positive.

. . .{-} . . . . (0) . . . . . . . . . . . (8) . . . . . . . . . . . (20) . . . . . . . . . .

Test x = 1:
M'(x) = 5x(20 - x)^2 (8 - x)
M'(1) = [positive][positive][positive] = positive.
Mark the region as positive.

. . .{-} . . . . (0) . . . . .{+} . . . . (8) . . . . . . . . . . . (20) . . . . . . . . . .

Test x = 10.
M'(x) = 5x(20 - x)^2 (8 - x)
M'(10) = [positive][positive][negative] = negative.

. . .{-} . . . . (0) . . . . .{+} . . . . (8) . . . . {-}. . . . . (20) . . . . . . . . . .

Test x = 21.
M'(x) = 5x(20 - x)^2 (8 - x)
M'(21) = [positive][positive][negative] = negative.

. . .{-} . . . . (0) . . . . .{+} . . . . (8) . . . . {-}. . . . . (20) . . . . {-}. . . .

In the negative regions, our function is decreasing; in the positive regions, the function is increase. That is

M(x) is increasing on [0, 8]
M(x) is decreasing on (-infinity, 0] U [8, infinity)

We're looking for a maximum, which occurs on the number line where we have an alternation from positive to negative. In this case, it occurs at x = 8.

From our first equation, since y = 20 - x, then y = 20 - 8, or
y = 12.

The two numbers are x = 8, y = 12. The actual maximum value is calculated by plugging in M(8).

M(8) = (8^2)(20 - 8)^3
M(8) = (64)(12)^3
M(8) = (64)(1728)
M(8) = 110592

************
I can tell you for a fact that 9 and 11 aren't your answer, because let's watch what happens when we plug them in.

(x^2)(y^3) = (9^2)(11^3) = (81)(1331) = 107811.
As you can see, this product is less than what we calculated above, 110592, so it can't possibly be the maximum.

Answer 3

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