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QuestionMonthly revenue is R(x)=79x−0.2x2R(x)=79x-0.2x^2 and cost is C(x)=30x+1600C(x)=30x+1600. Find wristwatches for max revenue and profit. Explain differences.

Studdy Solution

STEP 1

Assumptions1. The revenue function is given by R(x)=79x−0.xR(x)=79x-0.x^{} . The cost function is given by C(x)=30x+1600C(x)=30x+1600
3. The profit function is given by (x)=R(x)−C(x)(x)=R(x)-C(x)

STEP 2

To find the number of wristwatches that must be sold to maximize revenue, we need to find the maximum of the revenue function. This can be found by taking the derivative of the revenue function and setting it equal to zero.
R′(x)=0R'(x) =0

STEP 3

Calculate the derivative of the revenue function.
R′(x)=79−0.xR'(x) =79 -0.x

STEP 4

Set the derivative of the revenue function equal to zero and solve for xx.
79−0.4x=079 -0.4x =0

STEP 5

olve the equation for xx.
x=790.4x = \frac{79}{0.4}

STEP 6

Calculate the value of xx.
x=790.4=197.5x = \frac{79}{0.4} =197.5Since the number of wristwatches sold cannot be a fraction, we round xx to the nearest integer. So, the firm must sell198 wristwatches to maximize revenue.

STEP 7

Substitute x=198x =198 into the revenue function to find the maximum revenue.
R(198)=79⋅198−0.2⋅(198)2R(198) =79 \cdot198 -0.2 \cdot (198)^{2}

STEP 8

Calculate the maximum revenue.
R(198)=79⋅198−0.2⋅(198)2=$15,642.8R(198) =79 \cdot198 -0.2 \cdot (198)^{2} = \$15,642.8The maximum revenue is \$15,642.80.

STEP 9

The profit function is given by (x)=R(x)−C(x)(x)=R(x)-C(x). Substitute the given revenue and cost functions into the profit function.
(x)=(79x−.2x2)−(30x+160)(x) = (79x-.2x^{2}) - (30x+160)

STEP 10

implify the profit function.
(x)=49x−0.2x2−1600(x) =49x -0.2x^{2} -1600

STEP 11

To find the number of wristwatches that must be sold to maximize profit, we need to find the maximum of the profit function. This can be found by taking the derivative of the profit function and setting it equal to zero.
′(x)=0'(x) =0

STEP 12

Calculate the derivative of the profit function.
′(x)=49−0.4x'(x) =49 -0.4x

STEP 13

Set the derivative of the profit function equal to zero and solve for xx.
49−0.x=049 -0.x =0

STEP 14

olve the equation for xx.
x=490.4x = \frac{49}{0.4}

STEP 15

Calculate the value of xx.
x=490.4=122.5x = \frac{49}{0.4} =122.5Since the number of wristwatches sold cannot be a fraction, we round xx to the nearest integer. So, the firm must sell123 wristwatches to maximize profit.

STEP 16

Substitute x=123x =123 into the profit function to find the maximum profit.
(123)=49⋅123−0.2⋅(123)2−1600(123) =49 \cdot123 -0.2 \cdot (123)^{2} -1600

STEP 17

Calculate the maximum profit.
(123)=49⋅123−0.2⋅(123)2−1600=$,977.6(123) =49 \cdot123 -0.2 \cdot (123)^{2} -1600 = \$,977.6The maximum profit is \$,977.60.

STEP 18

The answers found in parts (a) and (c) differ because maximizing revenue and maximizing profit are two different objectives. The number of wristwatches that maximizes revenue does not necessarily maximize profit because the cost of producing and selling the wristwatches also needs to be taken into account when calculating profit.
A quadratic function is a reasonable model for revenue because it captures the idea that revenue increases with the number of wristwatches sold up to a point, but then decreases due to factors such as market saturation and decreased demand.

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