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Find the two positive numbers such that the sum of the first number squared and the second number is 57 and the product is a maximum.

The aim of this question is to learn how to maximize a certain function using the derivative approach.

In the derivative approach, we simply define the function that we want to maximize. Then we find the first derivative of this function and equate it to zero to find its roots. Once we have this value, we can check if it’s a maximum by plugging it into the second derivative through the second derivative test in case we have more than roots.

Expert Answer

Let x and y be the two numbers that we need to find. Now under the first constraint:

\[ x^2 \ + \ y \ = \ 57 \]

\[ y \ = \ 57 \ – \ x^2 \]

Under the second constraint, we need to maximize the following function:

\[ P(x,y) \ =\ xy \]

Substituting the value of y from the first constraint into the second one:

\[ P(x) \ =\ x ( 57 \ – \ x^2 ) \]

\[ P(x) \ =\ 57 x \ – \ x^3 \]

Taking the derivative of P(x):

\[ P'(x) \ =\ 57 \ – \ 3 x^2 \]

Equating first derivative to zero:

\[ 57 \ – \ 3 x^2 \ = \ 0\]

\[ 3 x^2 \ = \ 57 \]

\[ x \ = \ \sqrt{ \dfrac{ 57 }{ 3 } } \]

\[ x \ = \ \sqrt{ 19 } \]

\[ x \ = \ \pm 4.36 \]

Since we need positive number:

\[ x \ = \ + \ 4.36 \]

The second number y can be found by:

\[ y \ = \ 57 \ – \ x^2 \]

\[ y \ = \ 57 \ – \ ( 4.36 )^2 \]

\[ y \ = \ 57 \ – \ 19 \]

\[ y \ = \ 38 \]

Numerical Results

\[ x \ = \ 4.36 \]

\[ y \ = \ 38 \]

Example

Find two positive numbers such that their product is maximum while the sum of the square of one and the other number is equal to 27.

Let x and y be the two numbers that we need to find. Now under the first constraint:

\[ x^2 \ + \ y \ = \ 27 \]

\[ y \ = \ 27 \ – \ x^2 \]

Under the second constraint, we need to maximize the following function:

\[ P(x,y) \ =\ xy \]

Substituting the value of y from the first constraint into the second one:

\[ P(x) \ =\ x ( 27 \ – \ x^2 ) \]

\[ P(x) \ =\ 27 x \ – \ x^3 \]

Taking the derivative of P(x):

\[ P'(x) \ =\ 27 \ – \ 3 x^2 \]

Equating first derivative to zero:

\[ 27 \ – \ 3 x^2 \ = \ 0\]

\[ 3 x^2 \ = \ 27 \]

\[ x \ = \ \sqrt{ \dfrac{ 27 }{ 3 } } \]

\[ x \ = \ \sqrt{ 9 } \]

\[ x \ = \ \pm 3 \]

Since we need positive number:

\[ x \ = \ + \ 3 \]

The second number y can be found by:

\[ y \ = \ 27 \ – \ x^2 \]

\[ y \ = \ 27 \ – \ ( 3 )^2 \]

\[ y \ = \ 27 \ – \ 9 \]

\[ y \ = \ 18 \]

Hence, 18 and 3 are the two positive numbers.

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