Optimization Problems in Economics

Solved Problems

Solution.

Let \(x\) be the number of products in a lot. If \(x \le 5000,\) the unit price is \(100\,\$\) by condition. If \(x \gt 5000,\) the price is calculated by the formula

In the first case, for \(x \le 5000,\) the maximum income is reached at \(x = 5000.\) It is equal to

In the second case, for \(x \gt 5000,\) the income is defined by the following function:

Find the derivative:

By equating it to zero, we determine the critical point:

Note that the second derivative of the function \(R\left( x \right)\) is always negative:

So the critical point we found corresponds to the maximum of the function \(R\left( x \right).\) Thus, the company's income is maximized when selling lots of \(x = 12500\) units. The maximum income will be equal to

Solution.

Let \(x\) be the order size (in thousands). The revenue is described by the following function:

We rewrite the part of the function for \(x \gt 10:\)

Take the derivative and compute the critical points:

We have one critical point \(x = 30.\)

As

the point \(x = 30\) is a point of maximum, that is the company has the largest revenue if the order size is \(30,000.\)

Solution.

The profit function is written in the form

Differentiate \(P\left( x \right):\)

The critical point is equal to

The value \(x = \frac{{a - d}}{{2b}}\) corresponds to the maximum of the profit function as the second derivative is negative:

Calculate the price at \(x = \frac{{a - d}}{{2b}}:\)

Solution.

The tax rate \(r\) is calculated by the formula

We investigate it for extreme values. Find the derivative:

You can see that the function \(r\left( W \right)\) has three critical points, but since the coefficients \(b, c \gt 0,\) then a meaningful solution exists only in the following point:

When passing through this value the derivative changes sign from minus to plus. Consequently, the function \(r\left( W \right)\) reaches a minimum here, that is the tax rate will be the least at this income level.

Solution.

Let \(x\) be the number of deductions of \(50\,\$\) from the base price of \(2000\,\$.\) Then the unit price (if selling more than \(1000\) items a month) is \(2000 - 50x.\) The total amount of sold products for the month is \(1000 + 50x\) units. Then the total revenue is described by the expression

Differentiating the function \(R\left( x \right),\) we find the extremum point:

Note that the second derivative of the function \(R\left( x \right)\) is negative:

Therefore, \(x = 10\) is a maximum point. Consequently, the revenue will be highest when the number of deductions is equal to \(x = 10.\) The unit price in this case is

and the sales volume per month is

Accordingly, the maximum revenue of the company is given by

Solution.

Suppose that \(d\) is the number of deductions of \(\$5\) from the base price of \(\$100.\) Then the unit price is \(100 - 5d.\)

The total amount of products sold for the month is \(2000 + 200d.\)

Hence, the total revenue is given by

We differentiate the function \(R\left( d \right)\) to find its critical points:

Use the Second Derivative Test to identify the type of the critical point:

so \(d = 5\) is a point of maximum.

Compute the maximum possible revenue per month:

Solution.

According to the problem description, the variable part of expenses depends on the speed as follows:

where \(k\) is a coefficient of proportionality. Then the total cost per hour is expressed by the formula

In one hour, the ship travels a distance equal to \(v.\) Therefore, the cost per mile is given by

The resulting expression is a function of the speed \(v.\) We investigate its extreme values:

At this value of \(v,\) the function \({C_1} \left( v \right)\) reaches a minimum as the derivative changes sign from minus to plus when passing through this point. Hence, at this speed the cost per mile will be the lowest. The minimum value of the cost per mile is equal to

that is equal to twice the fixed costs.

Solution.

We can write the fuel cost in the form

where \(k\) is a coefficient of proportionality.

Hence, the total cost per hour \(C\) is written as

In one hour, the ship travels a distance numerically equal to \(v,\) so the cost per \(1\) mile \({C_1}\) is given by

Now we can explore the function \({C_1}\left( v \right)\) for extreme values:

Using the First Derivative Test, we can show that \(v = \sqrt[3]{{\frac{p}{{2k}}}}\) is a point of minimum.

Substitute this value of \(v\) and calculate the minimum cost per \(1\) mile:

Solution.

The conditions \(Q'\left( L \right) \gt 0,\;Q^{\prime\prime}\left( L \right) \lt 0\) reflect a decline in labor productivity with increasing the number of employees and are often observed in practice. Assuming that the labor costs are proportional to the amount of employees, we can write the following expression for the profit:

where \(p\) is the unit price, \(Q\left( L \right)\) is the production volume mentioned above, \(qL\) are the total labor cost, \(C\) are fixed costs that do not depend on the amount of workers.

In this formula, \(pQ\left( L \right)\) expresses the income of the company for a certain period. As a result, the profit \(P\) is a function of the number of employees \(P\left( L \right).\)

Investigate extreme values of this function. The first derivative is given by

The profit function \(P\left( L \right)\) has a critical point \({L^*}\) provided

Since the second derivative is negative for all possible values of \(L,\) the critical point is a maximum. Thus, there always exists an optimal number of employees \({L^*}\) in the given system, at which the company has a maximum profit.

Solution.

Let \(x\) be the distance from city \(A\) to the highway entrance \(S\). Then the shipping cost for \(1\) ton of goods over the railroad \(AS\) is \(px.\) The length of the highway section \(SC\) is determined by the Pythagorean theorem and is given by

Accordingly, the shipping cost over the highway section is equal to

Consequently, the total shipping cost for \(1\) ton of goods from point \(A\) to point \(C\) is described by the function

To investigate extreme values of the function \(Q\left( x \right),\) we find its derivative:

Equating the derivative to zero, we determine the critical value of \(x:\)

Let's see how the sign of the derivative changes at the critical point. In the expression for the derivative

the numerator of the fraction varies more with increasing or decreasing \(x\) than the denominator. Therefore, the derivative will be negative in the left neighborhood of the critical point and positive to the right. Thus, this point corresponds to the minimum of the function \(Q\left( x \right).\)

Note that the solution is valid only if \(q \gt p,\) i.e. when the unit shipping cost by truck exceeds the shipping cost by rail. This condition, however, is not a single constraint on the possible values of \(p\) and \(q.\) Exploring the range of values for the ratio \(\frac{q}{p},\) we consider two limiting cases:

If \(x = 0,\) we obtain the following solution:

This ratio of the shipping costs \(\frac{q}{p}\) is as minimum as possible. A this value of \(\frac{q}{p},\) it is necessary to build the highway along the diagonal \(AC.\)

The other limiting case corresponds to the solution \(x = a:\)

Here the ratio \(\frac{q}{p}\) tends to infinity. It is clear that in practice this ratio is finite, i.e. the optimal branching point will always be in the range \(0 \le x \lt a.\)