Finding the Domain and Range of a Function

NP1: A firm's total daily cost C depends on its daily output Q, and the relationship is given by the linear function:

C(Q) = 150 + 7Q 

The firm has a production capacity limit of 100 units per day, so the output Q can vary only between 0 and 100 units inclusive. What is the domain and of the cost function?

Solution:

The variable Q represents the number of units produced per day. Since the firm can produce a minimum of 0 units and a maximum of 100 units, the domain of the function is:

∴ Domain = {Q: 0 ≤ Q ≤ 100}

Next,

C(Q) = 150 + 7Q

Computing the minimum and maximum values of C:

When Q = 0:

C = 150 + 7(0) = 150

When Q = 100:

C = 150 + 7(100) = 150 + 700 = 850

So, as Q increases from 0 to 100, C increases linearly from 150 to 850.

∴ Range: {C:150 ≤ C ≤ 850}

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NP2: A firm's daily revenue R is a function of the number of units sold Q, given by:

$$$$

R(Q) = 20Q

The firm can sell a maximum of 200 units per day. Find the domain and range of the revenue function.

Solution:

Q = units sold per day
 ∴ Domain = {Q : 0 ≤ Q ≤ 200}

Computing the range:

R (0) = 20×0 = 0

R (200) = 20×200 = 4000

∴ Range = {R: 0 ≤ R ≤ 4000}

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NP3: A factory’s total cost $C(Q)$ to produce Q gadgets is:

C(Q) = 500 + 12Q

The maximum production is 250 gadgets per day. Find the domain and range.

Solution:

Q = gadgets produced per day

Domain = {Q: 0 ≤ Q ≤ 250}

Computing the range:

C (0) = 500

C (250) = 500 + 12×250 = 500 + 3000 = 3500

∴ Range = {C: 500 ≤ C ≤ 3500}

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NP4:  A firm’s profit $P(Q)$ in thousands of rupees is given by:

$$P(Q)=-2Q^2+40Q$$

where Q is the number of products sold, and the maximum is 10 units. Find the domain and range.

Solution:

Q = units sold

 ∴ Domain = {Q: 0 ≤ Q ≤ 10}

This is a quadratic function opening downward (since -2 < 0). Maximum occurs at:

$$Q = \frac{-b}{2a} = \frac{-40}{2(-2)} = 10$$

Computing the range:

P (0) = 0

P (10) = -2(100) + 40(10) = -200 + 400 = 200

So, range:
∴ Range = {P: 0 ≤ P ≤ 200}

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NP5: A factory’s total cost $C(Q)$ to produce Q gadgets is:

$$C(Q)=500+12Q$$

The maximum production is 250 gadgets per day. Find the domain and range.

Solution:

Q = gadgets produced per day
∴ Domain = {Q: 0 ≤ Q ≤ 250}

Computing the range:

C (0) = 500

C (250) = 500 + 12×250 = 500 + 3000 = 3500

∴ Range = {C: 500 ≤ C ≤ 3500}

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NP6: The price P (in ₹) of a product is related to the demand Q by:

$$P(Q)=200-5Q$$

The maximum possible demand is 30 units. Find the domain and range of the demand function.

Solution:

Q = quantity demanded
∴ Domain = {Q: 0 ≤ Q ≤ 30}

Range:

P (0) = 200

P (30) = 200 - 5×30 = 200 - 150 = 50

∴ Range = {P: 50 ≤ P ≤ 200}

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