# Calculus

## Applications of the Derivative # Optimization Problems in Economics

## Solved Problems

Click or tap a problem to see the solution.

### Example 9

A company sells its products at unit price of 100 $if the lot size does not exceed 5000 units. At higher volume the company gives a discount of 5$ for each additional thousand exceeding the level of 5000. Determine the order volume at which the company has the largest income.

### Example 16

When a ship is sailing, the total cost per hour includes a fixed part $$p$$ and the fuel cost $$q$$ that is proportional to the cube of its speed, that is $q = k{v^3}.$ Determine the minimum value of the total cost per $$1$$ mile.

### Example 17

Suppose that the production output $$Q$$ depends on the number of workers $$L$$ and this dependence is described by a function $$Q\left( L \right)$$ (Figure $$1$$). Show that if the derivatives satisfy the conditions $Q'\left( L \right) \gt 0, Q^{\prime\prime}\left( L \right) \lt 0,$ then there is an optimal number of workers $${L^*},$$ when the profit is maximized.

### Example 18

Two cities $$A$$ and $$B$$ are located at the distance of $$a\,\text{miles}$$ from each other and are connected by a straight railroad. For the transportation of goods from city $$A$$ to city $$C$$ located $$b\,\text{miles}$$ from the railroad it is necessary to build a highway adjacent to the railway line (Figure $$2$$). Find the point $$S$$ on the railroad where the highway should be built to provide the most cost-effective transportation of goods. The shipping cost is $$p\,\$$ per ton-mile by rail and $$q\,\$$ per ton-mile by truck.

### Example 9.

A company sells its products at unit price of $$100\,\$$ if the lot size does not exceed $$5000$$ units. At higher volume the company gives a discount of $$5\,\$$ for each additional thousand exceeding the level of $$5000.$$ Determine the order volume at which the company has the largest income.

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

$p\left( x \right) = 100 - 5 \cdot \frac{{x - 5000}}{{1000}} = 100 - 0,005x + 25 = 125 - 0,005x.$

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

$R_1 = 5000 \cdot 100 = 500000\;\left(\\right).$

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

${R_2} = R\left( x \right) = xp\left( x \right) = x\left( {125 - 0,005x} \right) = 125x - 0,005{x^2}\;\left(\\right).$

Find the derivative:

$R'\left( x \right) = \left( {125x - 0,005{x^2}} \right)^\prime = 125 - 0,01x.$

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

$R'\left( x \right) = 0,\;\; \Rightarrow 125 - 0,01x = 0,\;\; \Rightarrow x = \frac{{125}}{{0,01}} = 12500.$

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

$R^{\prime\prime}\left( x \right) = \left( {125 - 0,01x} \right)^\prime = - 0,01 \lt 0.$

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

${R_{\max }} = 125 \cdot 12500 - 0,005 \cdot {12500^2} = 781250\;\left(\\right).$

### Example 10.

A company sells wholesale lots of goods at a price of $$\20$$ per thousand for orders of $$10,000$$ or less, with the charge per thousand decreased by $$50$$ cents for each thousand above $$10,000.$$ Determine the order size at which the company has the maximum revenue.

Solution.

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

$R\left( x \right) = \left\{ {\begin{array}{*{20}{l}} {20x, \text{ if } x \le 10}\\ {200 + \left( {20 - 0.5\left( {x - 10} \right)} \right)\left( {x - 10} \right), \text{ if }x \gt 10} \end{array}} \right.$

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

${R_2}\left( x \right) = 200 + \left( {20 - 0.5\left( {x - 10} \right)} \right)\left( {x - 10} \right) = 200 + \left( {25 - 0.5x} \right)\left( {x - 10} \right) = 200 + 25x - 0.5{x^2} - 250 + 5x = 30x - 50 - 0.5{x^2}$

Take the derivative and compute the critical points:

$R^\prime\left( x \right) = {R_2^\prime}\left( x \right) = \left( {30x - 50 - 0.5{x^2}} \right)^\prime = 30 - x;$

We have one critical point $$x = 30.$$

As

$R^{\prime\prime}\left( x \right) = \left( {30 - x} \right)^\prime = - 1 \lt 0,$

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

### Example 11.

The demand function for a product is given by the linearly decreasing equation $p\left( x \right) = a - bx,$ and the total cost of producing $$x$$ units is expressed by the linearly increasing equation $C\left( x \right) = c + dx,$ where $$a,b,c,d$$ are positive numbers and $$a \gt d.$$ Find the price that maximizes the profit.

Solution.

The profit function is written in the form

$P\left( x \right) = R\left( x \right) - C\left( x \right) = xp\left( x \right) - C\left( x \right) = x\left( {a - bx} \right) - \left( {c + dx} \right) = ax - b{x^2} - c - dx = \left( {a - d} \right)x - b{x^2} - c.$

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

$P^\prime\left( x \right) = \left[ {\left( {a - d} \right)x - b{x^2} - c} \right]^\prime = a - d - 2bx;$

The critical point is equal to

$P^\prime\left( x \right) = 0,\;\; \Rightarrow a - d - 2bx = 0,\;\; \Rightarrow x = \frac{{a - d}}{{2b}}.$

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

$P^{\prime\prime}\left( x \right) = \left( {a - d - 2bx} \right)^\prime = - 2b \lt 0.$

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

$p = a - bx = a - b \cdot \frac{{a - d}}{{2b}} = \frac{{2ab - ab + bd}}{{2b}} = \frac{{ab + bd}}{{2b}} = \frac{{a + d}}{2}.$

### Example 12.

A certain country uses a progressive tax system. The amount of tax consists of a linear part proportional to the income and a nonlinear part depending on the income by a power law. The total amount of tax is determined by the formula $T\left( W \right) = aW + {\left( {bW + c} \right)^p},$ where $$W$$ is the income; $$p$$ is the exponent, $$a, b, c$$ are some positive numbers. At what level of income the tax rate will be minimal?

Solution.

The tax rate $$r$$ is calculated by the formula

$r\left( W \right) = \frac{{T\left( W \right)}}{W} = \frac{{aW + {{\left( {bW + c} \right)}^p}}}{W} = a + \frac{{{{\left( {bW + c} \right)}^p}}}{W}.$

We investigate it for extreme values. Find the derivative:

$r'\left( W \right) = \left[ {a + \frac{{{{\left( {bW + c} \right)}^p}}}{W}} \right]^\prime = \frac{{p{{\left( {bW + c} \right)}^{p - 1}} \cdot W - {{\left( {bW + c} \right)}^p} \cdot 1}}{{{W^2}}} = \frac{{{{\left( {bW + c} \right)}^{p - 1}}\left[ {pW - \left( {bW + c} \right)} \right]}}{{{W^2}}} = \frac{{{{\left( {bW + c} \right)}^{p - 1}}\left[ {\left( {p - b} \right)W - c} \right]}}{{{W^2}}}.$

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:

$\left( {p - b} \right)W - c = 0,\;\; \Rightarrow W = \frac{c}{{p - b}}.$

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.

### Example 13.

A company produces and sells $$1000$$ units per month at a price of $$2000\,\.$$ By reducing the price by $$50\,\$$ the company can additionally sell $$50$$ units per month. Determine the price at which the company has a maximum revenue and calculate this maximum value.

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

$R\left( x \right) = \left( {2000 - 50x} \right)\left( {1000 + 50x} \right) = 2000000 - 50000x + 100000x - 2500{x^2} = - 2500{x^2} + 50000x + 2000000\;\left(\\right).$

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

$R'\left( x \right) = \left( { - 2500{x^2} + 50000x + 2000000} \right)^\prime = - 5000x + 50000;$
$R'\left( x \right) = 0,\;\; \Rightarrow - 5000x + 50000 = 0,\;\; \Rightarrow x = \frac{{50000}}{{5000}} = 10.$

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

$R^{\prime\prime}\left( x \right) = \left( { - 5000x + 50000} \right)^\prime = - 5000 \lt 0.$

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

$2000 - 50 \cdot 10 = 1500\;\left(\\right),$

and the sales volume per month is

$1000 + 50 \cdot 10 = 1500\;\text{(units)}.$

Accordingly, the maximum revenue of the company is given by

${R_{\max }} = 1500 \cdot 1500 = 2,225,000\;\left(\\right)$

### Example 14.

A company produces and sells $$2000$$ units per month at a price of $$\ 100.$$ By reducing the price by $$\ 5$$ the company can additionally sell $$200$$ units per month. Determine the maximum possible revenue of the company.

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

$R\left( d \right) = \left( {100 - 5d} \right)\left( {2000 + 200d} \right) = 200000 - 10000d + 20000d - 1000{d^2} = 200000 + 10000d - 1000{d^2}.$

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

$R^\prime\left( d \right) = \left( {200000 + 10000d - 1000{d^2}} \right)^\prime = 10000 - 2000d;$
$R^\prime\left( d \right) = 0,\;\; \Rightarrow 10000 - 2000d = 0,\;\; \Rightarrow d = 5.$

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

$R^{\prime\prime}\left( d \right) = \left( {10000 - 2000d} \right)^\prime = - 2000 \lt 0,$

so $$d = 5$$ is a point of maximum.

Compute the maximum possible revenue per month:

$R\left( {d = 5} \right) = \left( {100 - 5 \cdot 5} \right)\left( {2000 + 200 \cdot 5} \right) = 75 \cdot 3000 = 225,000\left( \ \right).$

### Example 15.

When a ship is sailing, the fuel cost is proportional to the square of its speed relative to water. Besides that, there are fixed costs, which do not depend on the speed and are equal to $$p$$ (\$/hour). At what speed the total cost per $$1$$ mile will be the lowest?

Solution.

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

$q = k{v^2},$

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

$C = p + q = p + k{v^2}.$

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

${C_1} = {C_1}\left( v \right) = \frac{C}{v} = \frac{{p + k{v^2}}}{v} = \frac{p}{v} + kv.$

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

${C'_1} \left( v \right) = {\left( {\frac{p}{v} + kv} \right)^\prime } = - \frac{p}{{{v^2}}} + k = \frac{{k{v^2} - p}}{{{v^2}}};$
${C'_1} \left( v \right) = 0,\;\; \Rightarrow \frac{{k{v^2} - p}}{{{v^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {k{v^2} - p = 0}\\ {{v^2} \ne 0} \end{array},\;\; \Rightarrow v = \sqrt {\frac{p}{k}} } \right..$

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

$C = p + k{v^2} = p + k{\left( {\sqrt {\frac{p}{k}} } \right)^2} = 2p,$

that is equal to twice the fixed costs.

### Example 16.

When a ship is sailing, the total cost per hour includes a fixed part $$p$$ and the fuel cost $$q$$ that is proportional to the cube of its speed, that is $q = k{v^3}.$ Determine the minimum value of the total cost per $$1$$ mile.

Solution.

We can write the fuel cost in the form

$q = k{v^3},$

where $$k$$ is a coefficient of proportionality.

Hence, the total cost per hour $$C$$ is written as

$C = p + q = p + k{v^3}.$

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

${C_1} = {C_1}\left( v \right) = \frac{C}{v} = \frac{{p + k{v^3}}}{v} = \frac{p}{v} + k{v^2}.$

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

${C_1^\prime}\left( v \right) = \left( {\frac{p}{v} + k{v^2}} \right)^\prime = - \frac{p}{{{v^2}}} + 2kv = \frac{{2k{v^3} - p}}{{{v^2}}};$
${C_1^\prime}\left( v \right) = 0,\;\; \Rightarrow \frac{{2k{v^3} - p}}{{{v^2}}} = 0,\;\; \Rightarrow 2k{v^3} - p = 0,\;\; \Rightarrow {v^3} = \frac{p}{{2k}},\;\; \Rightarrow v = \sqrt{{\frac{p}{{2k}}}}.$

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

Substitute this value of $$v$$ and calculate the minimum cost per $$1$$ mile:

${C_{{1_{\min }}}} = {C_1}\left( {v = \sqrt{{\frac{p}{{2k}}}}} \right) = \frac{p}{{\sqrt{{\frac{p}{{2k}}}}}} + k{\left( {\sqrt{{\frac{p}{{2k}}}}} \right)^2} = \sqrt{{2k{p^2}}} + \sqrt{{\frac{{k{p^2}}}{4}}} = \sqrt{{k{p^2}}}\left( {\sqrt{2} + \frac{1}{{\sqrt{4}}}} \right) = \sqrt{{k{p^2}}}\left( {\sqrt{2} + \frac{1}{{\sqrt{4}}}} \right) = \sqrt{{k{p^2}}} \cdot \frac{{\sqrt{8} + 1}}{{\sqrt{4}}} = \sqrt{{k{p^2}}} \cdot \frac{3}{{\sqrt{4}}} = \sqrt{{k{p^2}}} \cdot \sqrt{{\frac{{27}}{4}}} = \sqrt{{\frac{{27k{p^2}}}{4}}}.$

### Example 17.

Suppose that the production output $$Q$$ depends on the number of workers $$L$$ and this dependence is described by a function $$Q\left( L \right)$$ (Figure $$1$$). Show that if the derivatives satisfy the conditions $Q'\left( L \right) \gt 0, Q^{\prime\prime}\left( L \right) \lt 0,$ then there is an optimal number of workers $${L^*},$$ when the profit is maximized.

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:

$P = P\left( L \right) = pQ\left( L \right) - qL - C,$

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

$P'\left( L \right) = \left[ {pQ\left( L \right) - qL - C} \right]^\prime = pQ'\left( L \right) - q.$

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

$pQ'\left( L^* \right) - q = 0,\;\; \Rightarrow Q'\left( L^* \right) = \frac{q}{p}.$

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.

### Example 18.

Two cities $$A$$ and $$B$$ are located at the distance of $$a\,\text{miles}$$ from each other and are connected by a straight railroad. For the transportation of goods from city $$A$$ to city $$C$$ located $$b\,\text{miles}$$ from the railroad it is necessary to build a highway adjacent to the railway line (Figure $$2$$). Find the point $$S$$ on the railroad where the highway should be built to provide the most cost-effective transportation of goods. The shipping cost is $$p\,\$$ per ton-mile by rail and $$q\,\$$ per ton-mile by truck.

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

$\left| {SC} \right| = \sqrt {{{\left( {a - x} \right)}^2} + {b^2}} .$

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

$q\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} .$

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

$Q = Q\left( x \right) = px + q\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} .$

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

$Q'\left( x \right) = \left( {px + q\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} } \right)^\prime = p + \frac{{q \cdot \left( { - 2} \right)\left( {a - x} \right)}}{{2\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} }} = p - \frac{{q\left( {a - x} \right)}}{{\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} }}.$

Equating the derivative to zero, we determine the critical value of $$x:$$

$Q'\left( x \right) = 0,\;\; \Rightarrow p - \frac{{q\left( {a - x} \right)}}{{\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} }} = 0,\;\; \Rightarrow p\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} = q\left( {a - x} \right),\;\; \Rightarrow {p^2}{\left( {a - x} \right)^2} + {p^2}{b^2} = {q^2}{\left( {a - x} \right)^2},\;\; \Rightarrow \left( {{q^2} - {p^2}} \right){\left( {a - x} \right)^2} = {p^2}{b^2},\;\; \Rightarrow a - x = \frac{{pb}}{{\sqrt {{q^2} - {p^2}} }},\;\; \Rightarrow x = a - \frac{{pb}}{{\sqrt {{q^2} - {p^2}} }} = a - \frac{b}{{\sqrt {{{\left( {\frac{q}{p}} \right)}^2} - 1} }}.$

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

$Q'\left( x \right) = p - \frac{{q\left( {a - x} \right)}}{{\sqrt {{{\left( {a - x} \right)}^2} + {b^2}} }} ,$

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:

$a - \frac{b}{{\sqrt {{{\left( {\frac{q}{p}} \right)}^2} - 1} }} = 0,\;\; \Rightarrow \frac{{a\sqrt {{{\left( {\frac{q}{p}} \right)}^2} - 1} - b}}{{\sqrt {{{\left( {\frac{q}{p}} \right)}^2} - 1} }} = 0,\;\; \Rightarrow a\sqrt {{{\left( {\frac{q}{p}} \right)}^2} - 1} = b,\;\; \Rightarrow {\left( {\frac{q}{p}} \right)^2} - 1 = {\left( {\frac{b}{a}} \right)^2},\;\; \Rightarrow \frac{q}{p} = \sqrt {{{\left( {\frac{b}{a}} \right)}^2} + 1} .$

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 $$x = a:$$

$a - \frac{b}{{\sqrt {{{\left( {\frac{q}{p}} \right)}^2} - 1} }} = a,\;\; \Rightarrow \frac{q}{p} \to \infty .$

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 $$0 \le x \lt a.$$