APPLICATION OF DERIVATIVES IN OPTIMIZATION IN BUSINESS

This post presents some examples of how the derivatives are applied in optimization in business.

Example 1: (Minimizing Average Cost)

In the making of some product, the total cost is TC = $(0.4Q² + 500Q + 16000). How many units must be produced to result in the minimum average cost (AC)? What is the minimum average cost?

Answer

The average cost (AC) is defined by $AC(Q) = \frac{TC}{Q}$ . Thus, in this case:

$AC(Q) = \frac{0.4 Q^2 + 500 Q + 16000}{Q}$

$AC(Q) = 0.4 Q + 500 + \frac{16000}{Q}$ ………………………………………………………………………………………………………………………………….. (*1)

To minimize AC, we have to solve the equation $\frac{dAC}{dQ} = 0$ . The derivative of AC with respect to Q is $\frac{dAC}{dQ} = 0.4 - 16000 Q^{-2}$ .

Since $\frac{dAC}{dQ} = 0$ , we have the following.

0.4 Q² – 16000 = 0

Q² – 40000 = 0

(Q+200)(Q-200) = 0

Q₁ = -200 and Q₂ = 200 [Q = -200 or Q = 200]

Here, we obtain two stationary points, i.e. Q₁ = -200 and Q₂ = 200. To determine which of them yields the minimum average cost, first we have to determine AC”(Q) = $\frac{d^2 AC }{dQ^2}$ , that is, the derivative of $\frac{dAC}{dQ}$ with respect to Q.

$\frac{d^2 AC}{dQ^2} = 32000 Q^{-3}$

If Q = -200 then the second derivative of AC with respect to Q is $\frac{d^2 AC}{dQ^2} = 32000 (-200)^{-3}$

Since the second derivative of AC with respect to Q is negative at that point, we can conclude that the value Q = -200 yields the relative maximum value of the average cost.

If Q = 200 then the value of the second derivative of AC with respect to Q is:

$\frac{d^2 AC}{dQ^2} = 32000 \cdot 200^{-3}$

Since the second derivative of AC with respect to Q is positive at that point, the value Q = 200 gives the relative minimum value of the average cost.

In conclusion, to minimize the average cost, the number of units that must be produced is 200.

To determine the minimum average cost, substitute Q = 200 into (*1). It follows that $AC(200) = 0.4 \cdot 200 + 500 + \frac{16000}{200} = 660$

Thus, the minimum average cost is $660/unit.

Example 2 (Maximizing Total Revenue)

The demand for a product satisfies the following equation: P = $(2700 – 9Q²)/unit. How many units must be sold to maximize the total revenue (TR)?

Answer

If Q units of the product are sold, the total revenue is TR = P⋅Q. Since P = 2700 – 9Q², we have:

TR(Q) = (2700 – 9Q²)⋅Q

TR(Q) = 2700Q – 9Q³ ……………………………………………………………………………………………………………………………………………………… ( *2)

To maximize TR, we have to solve the equation $\frac{dTR}{dQ} = 0$ . The derivative of TR with respect to Q is $\frac{dTR}{dQ} = 2700 - 27 Q^2$ .

As $\frac{dTR}{dQ} = 0$ , we get the following.

2700 – 27Q² = 0

100 – Q² = 0

(10 + Q)(10 – Q) = 0

Q₁ = -10 and Q₂ = 10 [Q = -10 or Q = 10]

Here, we obtain two stationary points, namely Q₁ = -10 and Q₂ = 10. To determine which of these values results in the maximum TR, we have to determine TR”(Q) = $\frac{d^2 TR }{dQ^2}$ , i.e. the derivative of $\frac{dTR}{dQ}$ with respect to Q.

$\frac{d^2 TR}{dQ^2} = - 54 Q$

If Q = -10 then the value of the second derivative of TR with respect to Q is $\frac{d^2 TR}{dQ^2} = -54.(-10) = 540$

Since the second derivative of TR with respect to Q is positive at that point, the value Q = -10 produces a relative minimum value of the total revenue.

If Q = 10 then the value of the second derivative of TR with respect to Q is $\frac{d^2 TR}{dQ^2} = -54 \cdot 10 = -540$

Since the second derivative of TR with respect to Q is negative at that point, the value Q = 10 gives a relative maximum value of the total revenue.

Thus, in order to maximize the total revenue, the number of units that must be sold is 10 units.

To determine the maximum total revenue, substitute Q = 10 into (*2). This results in the following.

TR(10) = 2700⋅10 – 9.10³ = 27000 – 9000 = 18000

So, the maximum total revenue is $18,000.

Example 3 (Maximizing Profit)

The demand for a product satisfies the equation P = $(90 – 3Q)/unit and the total cost (TC) to produce Q units of the product is TC = $(0.1 Q³ – 3Q² + 60Q + 100). How many products must be sold to gain the maximum profit? What is the maximum profit?

Answer

If Q units of the product are sold, the total revenue is TR = P⋅Q. Since P = 90 – 3Q, we get:

TR(Q) = (90 – 3Q)⋅Q

TR(Q) = 90Q – 3Q²

Furthermore, the profit function (π) is π = TR – TC.

π(Q) = (90Q – 3Q²) – (0.1 Q³ – 3Q² + 60Q + 100)

π(Q) = – 0.1 Q³ + 30Q – 100 ……………………………………………………………………………………………………………………………………………… (*3)

To maximize π, we have to solve the equation $\frac{d \pi}{dQ} = 0$ . It can be easily seen that $\frac{d \pi}{dQ} = - \frac{3}{10} Q^2 + 30$ .

Since $\frac{d \pi}{dQ} = 0$ , we get the following equation.

$- \frac{3}{10} Q^2 + 30 = 0$

Q² – 100 = 0

(Q + 10)(Q – 10) = 0

Q₁ = -10 and Q₂ = 10 [Q = -10 or Q = 10]

Here, we get two stationary points, namely Q₁ = -10 and Q₂ = 10. To determine which of them results in the maximum π, we compute π”(Q) = $\frac{d^2 \pi}{dQ^2}$ , that is the derivative of $\frac{d \pi}{dQ}$ with respect to Q.

$\frac{d^2 \pi}{dQ^2} = - \frac{3}{5} Q = - 0.6 Q$

If Q = -10 then the value of the second derivative of π with respect to Q is $\frac{d^2 \pi}{dQ^2} = - 0.6.(-10) = 6$

Since the second derivative of π with respect to Q is positive at that point, the value Q = -10 yields a relative minimum value of the profit.

If Q = 10 then the value of the second derivative of π with respect to Q is $\frac{d^2 \pi}{dQ^2} = - 0.6 \cdot 10 = - 6$

Since the second derivative of π with respect to Q is negative at that point, the value Q = 10 yields a relative maximum value of the profit.

Thus, in order to maximize the profit, the number of units that must be sold is 10.

To determine the maximum profit, substitute Q = 10 into (*3). Consequently, we get:

π(10) = – 0.1⋅10³ + 30⋅10 – 100 = 100

In conclusion, the maximum profit is $100.

On 25/03/2022

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