We want to find two numbers that meet a given restriction and then maximize their product.
The two numbers we want to get are:
[tex]A = \sqrt{19} \\B = 38[/tex]
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First, we need to see which is the restriction.
We know that the sum of the first number squared and the second number is 57, and we also know that the two numbers are positive.
Then if we define A as the first number and B as the second number, we have:
[tex]A^2 + B = 57[/tex]
Now we want to maximize their product, which can be written as:
[tex]P = A*B[/tex]
From the restriction, we can isolate B to get:
[tex]B = 57 - A^2[/tex]
Now we can replace it in the product equation:
[tex]P = A*(57 - A^2) = 57*A - A^3[/tex]
So now we have a polynomial of degree 3, and we want to maximize it.
To do this, we can try to find the zeros of the first derivative of P, which tells us where the maximums/minimums are located.
[tex]P' = 57 - 3*A^2[/tex]
So we need to solve:
[tex]0 = 57 - 3*A^2\\\\3*A^2 = 57\\\\A^2 = 57/3 = 19\\\\A = \pm\sqrt{19}[/tex]
Now remember that our numbers are positive, so we chose the positive option:
[tex]A = \sqrt{19}[/tex]
And from the equation:
[tex]B = 57 - A^2[/tex]
We can find the value of B.
[tex]B = 57 - \sqrt{19}^2 = 38[/tex]
Then the two numbers are:
[tex]A = \sqrt{19} \\B = 38[/tex]
And the product is:
[tex]P = \sqrt{19}*38 = 165.6[/tex]
The graph of
[tex]P = A*(57 - A^2) = 57*A - A^3[/tex]
can be seen below, where you can see where the maximum is at.
If you want to learn more, you can read:
https://brainly.com/question/11510063
