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.**