The aim of this question is to **understand** the solution of **word problems** related to simple **algebraic expressions** and the solution of a simple **system of linear equations**, and also the concept of **maximizing or minimizing** a given equation.

To solve such word problems one has to simply **convert the given constraints** and conditions into one or more **algebraic equations** in one or more variables. to find a **unique solution**, the **number of unknowns** must be **equal to** the no. of consistent or independent, or **unique algebraic equations**.

Once we have these equations, any **method of solving linear equations** or a system of linear equations may be deployed to find the unknown variables. Some well-known techniques include the **substitution**, **echelon form** of matrices, **Crammer’s rule**, etc.

To **maximize** the functions, we can deploy the **differentiation method** where we find the **roots of the equation** $ f^{ ‘ } ( x ) \ = \ 0 $.

## Expert Answer

Let $ x $ and $ y $ be the **two required positive real numbers**. Under the given conditions and constraints:

\[ x \ + \ y \ = \ 110 \]

\[ y \ = \ 110 \ – \ x \ … \ …. \ … \ ( 1 ) \]

Now the **product** of $ x $ and $ y $ is given by the **following formula**:

\[ x y \ = \ x ( 110 \ – \ x ) \]

\[ x y \ = \ 110 x \ – \ x^{ 2 } \]

Since we need to **maximize the product**, let’s call it $ f( x ) $:

\[ f ( x ) \ = \ 110 x \ – \ x^{ 2 } \]

**Differentiating both sides:**

\[ f^{ ‘ } ( x ) \ = \ 110 \ – \ 2 x \]

**Differentiating both sides:**

\[ f^{ ” } ( x ) \ = \ – 2 \]

Since $ f^{ ” } ( x ) < 2 $, so the **maxima exists at** $ f^{ ‘ } ( x ) \ = \ 0 $:

\[ 110 \ – \ 2 x \ = \ 0 \]

\[ 110 \ = \ 2 x \]

\[ x \ = \ \dfrac{ 110 }{ 2 } \]

\[ x \ = \ 55 \]

**Substituting this value in equation (1):**

\[ y \ = \ 110 \ – \ ( 55 ) \]

\[ y \ = \ 55 \]

So the **two numbers are** $ 55 $ and $ 55 $.

## Numerical Result

\[ x \ = \ 55 \]

\[ y \ = \ 55 \]

## Example

If two numbers’ **sum is equal to 600**, **maximize their product**.

Let $ x $ and $ y $ be the **two required positive real numbers**. Under the given conditions and constraints:

\[ x \ + \ y \ = \ 600 \]

\[ y \ = \ 600 \ – \ x \ … \ …. \ … \ ( 2 ) \]

Now the **product** of $ x $ and $ y $ is given by the **following formula**:

\[ x y \ = \ x ( 600 \ – \ x ) \]

\[ x y \ = \ 600 x \ – \ x^{ 2 } \]

Since we need to **maximize the product**, let’s call it $ f( x ) $:

\[ f ( x ) \ = \ 600 x \ – \ x^{ 2 } \]

**Differentiating both sides:**

\[ f^{ ‘ } ( x ) \ = \ 600 \ – \ 2 x \]

**Differentiating both sides:**

\[ f^{ ” } ( x ) \ = \ – 2 \]

Since $ f^{ ” } ( x ) < 2 $, so the **maxima exists at** $ f^{ ‘ } ( x ) \ = \ 0 $:

\[ 600 \ – \ 2 x \ = \ 0 \]

\[ 600 \ = \ 2 x \]

\[ x \ = \ \dfrac{ 600 }{ 2 } \]

\[ x \ = \ 300 \]

**Substituting this value in equation (1):**

\[ y \ = \ 600 \ – \ ( 300 ) \]

\[ y \ = \ 300 \]

So the **two numbers are** $ 300 $ and $ 300 $.