Optimization Problems Involving Numbers

Number problems involve finding two numbers that satisfy certain conditions.

If we label the numbers using the variables x and y, we can compose the objective function F (x, y) to be maximized or minimized.

The constraint specified in the problem allows to eliminate one of the variables.

When we get the objective function as a single variable function, we can use differentiation to find the extreme values.

Solved Problems

Solution.

Let $$ and $$ be the two numbers. We want to find the minimum of the function

As $$ we can eliminate one variable in the objective function. Substitute $$ in the objective function.

Take the derivative:

The critical points are

Note that the second derivative is positive:

Hence, the objective function has a local minimum at $$

So the sum of squares is a minimum when $$ $$

Solution.

Let $$ and $$ be the two numbers. The objective function is written in the form

As $$ we can substitute $$ into the objective function:

Take the derivative and find the critical points:

We should take only positive root $$

Find the second derivative and determine its sign at this point:

We see that $$ is a point of minimum by the Second Derivative Test.

Hence, the answer is

Solution.

The objective function is

where $$ and $$ are the two numbers.

Since $$ we can substitute $$ in the objective function above. This yields:

Take the derivative:

There is one critical point $$.

Note that the second derivative is always positive:

Hence, the objective function has a minimum at the point $$ The other number equals $$

Solution.

The objective function is given by

where $$ are the two unknown numbers.

As $$ we obtain:

The derivative of the objective function is

Now it is easy to find the critical points:

so the critical points are

We should take only the point $$ Then $$

Solution.

The function to be minimized is written as

where $$ is supposed to be a positive number.

Take the derivative:

There are the following critical values:

Only the root $$ satisfies the condition $$

Determine the second derivative:

As the second derivative is positive at $$ this point corresponds to the minimum of the objective function. The minimum value of the function is

Solution.

Let $$ and $$ be the two numbers. The objective function is written as

As $$ we substitute $$ in the function above:

Compute the derivative:

so the critical point is $$

The second derivative is

Hence, $$ corresponds to the minimum of the objective function by the Second Derivative Test. The other number equals $$