Math 018

Elementary Linear Algebra

Spring 2011

Homework 3 - Solutions Problem 1. Let f (x) = 7 − 5x and g(x) = 2x − 3. Find the following. (a) f (2) (b) g(−2) (c) g(k 2 ) (d) f (g(0)) Solution.

(a) f (x) = 7 − 5x f (2) = 7 − 5 × 2 = 7 − 10 f (2) = −3 .

=⇒ =⇒

(b) =⇒ =⇒

g(x) = 2x − 3 g(−2) = 2(−2) − 3 = −4 − 3 g(−2) = −7 .

(c) =⇒

g(x) = 2x − 3 g(k 2 ) = 2k 2 − 3 .

(d) =⇒

g(x) = 2x − 3 g(0) = 2(0) − 3 = −3.

So, f (g(0)) = f (−3). Now, =⇒ =⇒

f (x) = 7 − 5x f (−3) = 7 − 5(−3) = 7 + 15 f (−3) = 22.

Thus, f (g(0)) = 22.

Problem 2. Suppose the fixed cost for producing an item is $100, and 50 units of the item cost $1000 to produce. Find the cost function, given that it is linear. (Define any variable that needs to be used.)

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Math 018

Elementary Linear Algebra

Spring 2011

Solution. Let m be the marginal cost, that is, the cost of production per unit of the item, and x denote the number of units of the item produced. Then since the fixed cost for producing the item is $100, the linear cost function, C(x) is given by C(x) = mx + 100. We know that 50 units of the item cost $1000 to produce. So, =⇒ =⇒ =⇒

C(50) = 1000 m(50) + 100 = 1000 50m = 900 900 = 18. m= 50

Thus, the cost function is C(x) = 18x + 100. Problem 3. Write a linear cost function for the following situation. (Define any variable that needs to be used.) “For a one-day rental, a car rental firm charges $44 plus 28 cents per mile.” Solution. From the problem statement, the fixed cost of a one-day rental is $44. Let x denote the number of miles traveled. Then the linear cost function, C(x) which represents the cost of traveling x miles in a day on the rental car, is given by C(x) = 0.28x + 44.

Problem 4. Let the supply and demand functions for sugar be given by p = S(q) = 1.4q − 0.6 p = D(q) = −2q + 3.2,

and

where p is the price per pound and q is the quantity in thousands of pounds. (a) Find the supply and demand at a price of 80 cents per pound. (b) Find the equilibrium quantity. (c) Find the equilibrium price. Solution.

(a) Setting p = 0.80 in the supply equation, we have =⇒ =⇒ =⇒

0.8 = S(q) = 1.4q − 0.6 0.8 = 1.4q − 0.6 1.4q = 0.8 + 0.6 = 1.4 q = 1.

This means that the supply at a price of 80 cents per pound is 1000 pounds. Penn State University, University Park

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Math 018

Elementary Linear Algebra

Spring 2011

Similarly, setting p = 0.80 in the demand equation, we have =⇒ =⇒ =⇒

0.8 = D(q) = −2q + 3.2 0.8 = −2q + 3.2 2q = 3.2 − 0.8 = 2.4 2.4 = 1.2. q= 2

This means that the demand at a price of 80 cents per pound is 1200 pounds.

(b) At the equilibrium point, the demand is equal to the supply. So, D(q) = S(q) −2q + 3.2 = 1.4q − 0.6 2q + 1.4q = 3.2 + 0.6 3.4q = 3.8 38 19 3.8 = = ≈ 1.11765. q= 3.4 34 17

=⇒ =⇒ =⇒ =⇒

Thus the equilibrium quantity is approximately 1117.65 pounds.

(c) To find the equilibrium price, we need to substitute the equilibrium quantity for q into either D(q) or S(q).     19 19 = −2 + 3.2 ≈ 0.96, and D 17 17  S

19 17



 = 1.4

19 17

 − 0.6 ≈ 0.96 also, as it should be.

Thus the equilibrium price is approximately 96 cents per pound.

Problem 5. Let the supply and demand functions for butter pecan ice cream be given by p = S(q)

=

2 q 5

and

2 p = D(q) = 100 − q, 5 where p is the price in dollars and q is the number of 10-gallon tubs. (a) Find the supply and demand at a price of $10. (b) Find the equilibrium quantity. Penn State University, University Park

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Math 018

Elementary Linear Algebra

Spring 2011

(c) Find the equilibrium price. Solution.

(a) Setting p = 10 in the supply equation, we have

=⇒ =⇒

2 10 = S(q) = q 5 2q = 5 × 10 50 q= = 25. 2

Thus, the supply at a price of $10 per 10-gallon tub is 25 tubs. Similarly, setting p = 10 in the demand equation, we have

=⇒ =⇒ =⇒

2 10 = D(q) = 100 − q 5 2 q = 90 5 2q = 90 × 5 450 = 225. q= 2

Thus, the demand at a price of $10 per 10-gallon tub is 225 tubs.

(b) At the equilibrium point, the supply is equal to the demand, that is,

=⇒ =⇒ =⇒

D(q) = S(q) 2 2 100 − q = q 5 5 4 q = 100 5 100 × 5 q= = 125. 4

Thus, the equilibrium quantity is 125 10-gallon tubs.

(c) To find the equilibrium price, we need to substitute the equilibrium quantity for q into either D(q) or S(q). 2 D(125) = 100 − × 125 = 100 − 50 = 50, 5 2 S(125) = × 125 = 50 also, as expected. 5

and

Thus, the equilibrium price is $50.

Problem 6. The cost of producing x units of a religious medal is given by C(x) = 12x + 39. The revenue is R(x) = 25x. Both C(x) and R(x) are in dollars. Penn State University, University Park

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Math 018

Elementary Linear Algebra

Spring 2011

(a) Find the break-even quantity. (b) Find the profit from 250 units. (c) Find the number of units that must be produced for a profit of $130. Solution.

(a) At the break-even point, the cost is equal to the revenue, that is, =⇒ =⇒ =⇒ =⇒

C(x) = R(x) 12x + 39 = 25x 25x − 12x = 39 13x = 39 39 = 3. x= 13

Thus, the break-even quantity is 3 units of the religious medal.

(b) The profit function, P (x) is given by =⇒ =⇒

P (x) = R(x) − C(x) P (x) = 25x − (12x + 39) P (x) = 13x − 39.

Thus, the profit from 250 units is P (250) = 13(250) − 39 = 3250 − 39 = 3211. Profit from 250 units = $3211.

(c) Profit = $130 implies =⇒ =⇒ =⇒

P (x) = 130 13x − 39 = 130 13x = 130 + 39 = 169 169 x= = 13. 13

Thus, 13 units of the religious medal need to be produced for a profit of $130.

Problem 7. The fixed cost for producing tacos is $20 and the marginal cost is $5 per taco. The tacos are to be sold at $15 each. Let x denote the number of tacos. Find the following. (a) The cost function, C(x). (b) The revenue function, R(x). (c) The profit function, P (x). Penn State University, University Park

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Math 018

Elementary Linear Algebra

Spring 2011

(d) The break-even quantity. (e) The profit from selling 100 tacos. (f) The number of tacos that need to be sold to produce a profit of $500. Solution.

(a) A linear cost function, C(x) is given by C(x) = mx + b,

where m is the marginal cost, and b is the fixed cost. The fixed cost for producing tacos is $20, and the marginal cost is $5 per taco. Thus, the cost of producing x tacos is given by C(x) = 5x + 20.

(b) The revenue function, R(x) is given by R(x) = px, where p is the selling price per unit of the item and x is the number of units. Since the tacos sell for $15 each, the revenue function is R(x) = 15x.

(c) The profit function, P (x) is given by P (x) = R(x) − C(x), where R(x) is the revenue from selling x units and C(x) is the cost of producing x units. Using the expressions for C(x) and R(x) found in the previous two questions, we have P (x) = 15x − (5x + 20) =⇒ P (x) = 10x − 20.

(d) At the break-even point, the revenue equals cost. So, =⇒ =⇒ =⇒ =⇒

C(x) = R(x) 5x + 20 = 15x 15x − 5x = 20 10x = 20 x = 2.

Thus, the break-even quantity is 2 tacos.

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Math 018

Elementary Linear Algebra

Spring 2011

(e) The profit from selling x tacos is given by P (x) = 10x − 20. Thus, the profit from selling 100 tacos is P (100) = 10(100) − 20 = 1000 − 20 = 980. The profit from selling 100 tacos is $980.

(f) A profit of $500 implies =⇒ =⇒ =⇒

P (x) = 500 10x − 20 = 500 10x = 520 x = 52.

Thus, 52 tacos need to be sold to produce a profit of $500.

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