# Get the Knowledge that sets you free...Science and Math for K8 to K12 students

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## Applications of Derivatives

Rate of Change

If we are pumping gas into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. We may wish to study the relationship among these parameters. For example, if we know that the volume of balloon is increasing at a certain rate, then we may wish to know how that affects the rate of change of the radius. This consideration gives rise to related rates problems. Here is a general procedure for solving related rates problems.

General procedure for solving related rates problems
Step 1: Read the given problem carefully and, if appropriate, draw a diagram.
Step 2: Represent the given information and unknowns by mathematical symbols.
Step 3: Write an equation involving the rate of change to be determined. If the equation contains more than one variable, it may be necessary to reduce the equation to one variable.
Step 4: Differentiate each term of the equation which is obtained in above step with respect to time.
Step 5: Substitute all the known rates of change and known values into the equation which is obtained in the above step, and solve it for the desired rate of change.
Step 6: Write the answer and indicate the units of measure.
• Ex 1: If a snowball melts so that its surface area decreases at a rate of 1 cm2/min, then find the rate at which the diameter decreases when the diameter is 12 cm.

Sol:
 Step 1: Define the variables. Let 'A' be the surface area of the snowball; 'D' be the diameter of snowball; 't' be the time in minutes. Step 2: Given: = 1 cm2/min; Find: at D = 12 cm. Step 3: Set up an equation: A = πD2. Step 4: Differentiate both sides with respect to time 't'. Step 5: Substitute the known values in above equation. Step 6: Thus, the diameter of snowball decreases at the rate of 0.013 cm/min.
• Ex 2: A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 2 m/s, how fast is the boat approaching the dock when it is 10 m from the dock ?

Sol:
 Step 1: Define the variables. Let 'L' is the length of the rope; 'x' is the distance of the boat from the dock and 't' be the time in seconds. Step 2: Given: = 2 m/sec; Find: at x = 10 m. Step 3: Set up an equation: . Step 4: Differentiate both sides with respect to time 't'. Step 5: Substitute the known values in above equation. Step 6: Thus, the boat is approaching the dock at 2.01 m/s.
Optimization Problems

The methods we have learned in the previous chapter for finding extreme values have practical applications in most of the areas of life. In this section and the next we solve such problems as maximizing areas, volumes, profits and minimizing distances, costs, and times.

In solving such practical problems the greatest challenge often is to convert the word problem into a mathematical optimization problem by setting up the function that is to be maximized or minimized. Here is the general procedure for solving the optimization problem:

General procedure for solving optimization problems
Step 1: Read the given problem carefully and, if appropriate, draw a diagram.
Step 2: From the given problem, determine what is given and what is to be found and represent this information by mathematical symbols.
Step 3: Write an equation that is a function of the variable representing the quantity to be minimized or maximized.
Step 4: If the equation involves more than one variable, then reduce the equation to a single variable that represents the quantity to be minimized or maximized.
Step 5: Based on the information given in the problem, determine the appropriate interval or domain for the above equation.
Step 6: Differentiate the resultant function to obtain the first order derivative and to find critical numbers [i.e., the numbers at which first order derivative = '0' or does not exist].
Step 7: Apply the first derivative test or the second derivative test by finding the second order derivative.
Step 8: Check the function values at the endpoints of the interval. Finally, write the answer(s) to the given problem and, if given, indicate the units of measure.
• Ex: A rectangular plot of farmland will be bounded on 1 side by a river and on the other three sides by a single-strand electric fence. With 500 m of wire at our disposal, what is the largest area that we can enclose and what are its dimensions?

Sol:
Step 1: Draw a diagram.
Step 2: Let 'A' be the area of rectangular plot formed, 'y' be the length of the fence opposite the river and the other two sides each having length 'x'. The sum of these sides should be equal to the total length of the electric fence, i.e., x + x + y = 500 ⇒ 2x + y = 500.
Step 3: The area of rectangular plot formed is: A = xy.
Step 4: The equation A = xy involves two variables x and y. Therefore, we reduce the equation to a single variable by as follows:  A = xy = x(500 – 2x) [∵ 2x + y = 500] = 500x – 2x2
Step 5: Note that x ≥ 0 and x ≤ 250 (otherwise A < 0). So the function that we wish to maximize is: A(x) = 500x – 2x2    0 ≤ x ≤ 250.
Step 6: The derivative of A(x) is: A'(x) = 500 – 4x, so to find critical numbers we set A'(x) = 0. By solving, we get x = 125.
Step 7: The second order derivative of the above function is: A"(x) = –4 (< 0 for all 'x'). Since the second order derivative is negative, the function has relative maximum at x = 125.
Step 8: By substituting x = 125 in A(x) = 500x – 2x2, we get maximum area as:  A(125) = 500(125) – 2(125)2 = 62500 – 31250 = 31250 m2
Therefore, area that we can enclose is 31250 m2 and its dimensions are: 125 m by 250 m.
Applications to Economics

Cost function: The total cost 'C' of producing and marketing 'x' units of a product depends upon the number of units (x). So the function relating 'C' and 'x' is called "Cost function" and is written as "C = C(x)". The total cost of producing 'x' units of the product consists of two parts: (i) fixed cost and (ii) variable cost.

• Fixed cost: The fixed cost consists of all types of costs which do not change with the level of production. Ex: salaries, rent, insurance, utilities, etc.
• Variable cost: The variable cost is the sum of all costs that are dependent on the level of production. Ex: the cost of material, cost of packaging, etc.

Demand function: An equation that relates price per unit and quantity demanded at that price is called a demand function or price function. If 'p' is the price per unit of a certain product and 'x' is the number of units demanded, then 'p(x)' is called the demand function or price function.

Revenue function: If 'x' is the number of units of certain product sold at a rate of Rs. 'p' per unit, then the amount derived from the sale of 'x' units of a product is the total revenue. Thus, if 'R' represents the total revenue from 'x' units of the product at the rate of Rs. 'p' per unit, then R = [x.p] is the total revenue. Thus, if R(x) is the revenue function, then R(x) = x . p(x).

Profit function: The profit is calculated by subtracting the total cost from the total revenue obtained by selling 'x' units of a product. Thus, if P(x) is the profit function, then P(x) = R(x) – C(x).

Average cost function: Let C = C(x) be the total cost of producing and selling 'x' units of a certain commodity, then the average cost (AC) is the cost per unit, that is, AC = . Thus, if c(x) is the average cost function, then c(x) = .

Marginal cost function: Let C = C(x) be the total cost of producing 'x' units of a certain product, then the marginal cost (MC) is the rate of change of C with respect to 'x'. In other words, the marginal cost function is the derivative of the cost function C(x). Thus, MC = .

Average revenue function: If R is the total revenue obtained by selling 'x' units of a certain commodity at a price 'p' per unit, then the average revenue (AR) is the revenue per unit, that is, AR = .

Marginal revenue function: If R is the total revenue obtained by selling 'x' units of a certain commodity at a price 'p' per unit, then the marginal revenue (MR) is the rate of change of total revenue with respect to the quantity demanded (x). In other words, the marginal revenue function is the derivative, R'(x), of the revenue function R(x). Thus, MR = .

Points to remember
• If the average cost is minimum, then marginal cost = average cost.
• If the profit is a maximum, then marginal revenue = marginal cost.