QuestionGiven revenue $R(x)=-1.9 x^{2}+323 x$ and cost $C(x)=0.09 x^{3}-2 x^{2}+75 x+600$ :
(a) Find profit $P(x)=R(x)-C(x)$ .
(b) Calculate profit for $x=25$ hundred phones.
(c) Explain $P(25)$ .

Studdy Solution

STEP 1

Assumptions1. The revenue function is $R(x)=-1.9 x^{}+323 x$ . The cost function is $C(x)=0.09 x^{3}- x^{}+75 x+600$
3. The profit function is defined as $(x)=R(x)-C(x)$ 4. The value of $x$ is in hundreds of cell phones sold

STEP 2

We first need to find the profit function, $(x)$ . This is done by subtracting the cost function $C(x)$ from the revenue function $R(x)$ .
$(x) = R(x) - C(x)$

STEP 3

Substitute the given functions into the equation.
$(x) = (-1.9 x^{2}+323 x) - (0.09 x^{3}-2 x^{2}+75 x+600)$

STEP 4

istribute the minus sign to all the terms in the parentheses.
$(x) = -1.9 x^{2}+323 x -0.09 x^{3}+2 x^{2}-75 x-600$

STEP 5

Rearrange the terms in descending order of power.
$(x) = -0.09 x^{3}+0.1 x^{2}+248 x-600$ This is the profit function.

STEP 6

To find the profit when $x=25$ hundred cell phones are sold, substitute $x=25$ into the profit function.
$(25) = -0.09 (25)^{3}+0.1 (25)^{2}+248 (25)-600$

STEP 7

Calculate the value of $(25)$ .
$(25) = -0.09 (15625)+0.1 (625)+248 (25)-600$

STEP 8

Continue the calculation.
$(25) = -1406.25+62.5+6200-600$

STEP 9

Calculate the final value of $(25)$ .
$(25) = \$4256.25$ This is the profit when $x=25$ hundred cell phones are sold.

STEP 10

To interpret $(25)$ , we note that $(25)$ is the profit when $x=25$ hundred cell phones are sold. This means that when2500 cell phones are sold, the profit is \$4256.25.

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QuestionGiven revenue $R(x)=-1.9 x^{2}+323 x$ and cost $C(x)=0.09 x^{3}-2 x^{2}+75 x+600$ :
(a) Find profit $P(x)=R(x)-C(x)$ .
(b) Calculate profit for $x=25$ hundred phones.
(c) Explain $P(25)$ .

Studdy Solution

STEP 1

Assumptions1. The revenue function is $R(x)=-1.9 x^{}+323 x$ . The cost function is $C(x)=0.09 x^{3}- x^{}+75 x+600$
3. The profit function is defined as $(x)=R(x)-C(x)$ 4. The value of $x$ is in hundreds of cell phones sold

STEP 2

We first need to find the profit function, $(x)$ . This is done by subtracting the cost function $C(x)$ from the revenue function $R(x)$ .
$(x) = R(x) - C(x)$

STEP 3

Substitute the given functions into the equation.
$(x) = (-1.9 x^{2}+323 x) - (0.09 x^{3}-2 x^{2}+75 x+600)$

STEP 4

istribute the minus sign to all the terms in the parentheses.
$(x) = -1.9 x^{2}+323 x -0.09 x^{3}+2 x^{2}-75 x-600$

STEP 5

Rearrange the terms in descending order of power.
$(x) = -0.09 x^{3}+0.1 x^{2}+248 x-600$ This is the profit function.

STEP 6

To find the profit when $x=25$ hundred cell phones are sold, substitute $x=25$ into the profit function.
$(25) = -0.09 (25)^{3}+0.1 (25)^{2}+248 (25)-600$

STEP 7

Calculate the value of $(25)$ .
$(25) = -0.09 (15625)+0.1 (625)+248 (25)-600$

STEP 8

Continue the calculation.
$(25) = -1406.25+62.5+6200-600$

STEP 9

Calculate the final value of $(25)$ .
$(25) = \$4256.25$ This is the profit when $x=25$ hundred cell phones are sold.

STEP 10

To interpret $(25)$ , we note that $(25)$ is the profit when $x=25$ hundred cell phones are sold. This means that when2500 cell phones are sold, the profit is \$4256.25.

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Studdy Solution

STEP 1

STEP 2

We first need to find the profit function, (x)(x)(x). This is done by subtracting the cost function C(x)C(x)C(x) from the revenue function R(x)R(x)R(x).(x)=R(x)−C(x)(x) = R(x) - C(x)(x)=R(x)−C(x)

STEP 3

Substitute the given functions into the equation.(x)=(−1.9x2+323x)−(0.09x3−2x2+75x+600)(x) = (-1.9 x^{2}+323 x) - (0.09 x^{3}-2 x^{2}+75 x+600)(x)=(−1.9x2+323x)−(0.09x3−2x2+75x+600)

STEP 4

istribute the minus sign to all the terms in the parentheses.(x)=−1.9x2+323x−0.09x3+2x2−75x−600(x) = -1.9 x^{2}+323 x -0.09 x^{3}+2 x^{2}-75 x-600(x)=−1.9x2+323x−0.09x3+2x2−75x−600

STEP 5

Rearrange the terms in descending order of power.(x)=−0.09x3+0.1x2+248x−600(x) = -0.09 x^{3}+0.1 x^{2}+248 x-600(x)=−0.09x3+0.1x2+248x−600This is the profit function.

STEP 6

To find the profit when x=25x=25x=25 hundred cell phones are sold, substitute x=25x=25x=25 into the profit function.(25)=−0.09(25)3+0.1(25)2+248(25)−600(25) = -0.09 (25)^{3}+0.1 (25)^{2}+248 (25)-600(25)=−0.09(25)3+0.1(25)2+248(25)−600

STEP 7

Calculate the value of (25)(25)(25).(25)=−0.09(15625)+0.1(625)+248(25)−600(25) = -0.09 (15625)+0.1 (625)+248 (25)-600(25)=−0.09(15625)+0.1(625)+248(25)−600

STEP 8

Continue the calculation.(25)=−1406.25+62.5+6200−600(25) = -1406.25+62.5+6200-600(25)=−1406.25+62.5+6200−600

STEP 9

Calculate the final value of (25)(25)(25).(25)=$4256.25(25) = \$4256.25(25)=$4256.25This is the profit when x=25x=25x=25 hundred cell phones are sold.

STEP 10

To interpret (25)(25)(25), we note that (25)(25)(25) is the profit when x=25x=25x=25 hundred cell phones are sold. This means that when2500 cell phones are sold, the profit is \$4256.25.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

We first need to find the profit function, $(x)$ . This is done by subtracting the cost function $C(x)$ from the revenue function $R(x)$ .
$(x) = R(x) - C(x)$

Substitute the given functions into the equation.
$(x) = (-1.9 x^{2}+323 x) - (0.09 x^{3}-2 x^{2}+75 x+600)$

istribute the minus sign to all the terms in the parentheses.
$(x) = -1.9 x^{2}+323 x -0.09 x^{3}+2 x^{2}-75 x-600$

Rearrange the terms in descending order of power.
$(x) = -0.09 x^{3}+0.1 x^{2}+248 x-600$ This is the profit function.

To find the profit when $x=25$ hundred cell phones are sold, substitute $x=25$ into the profit function.
$(25) = -0.09 (25)^{3}+0.1 (25)^{2}+248 (25)-600$

Calculate the value of $(25)$ .
$(25) = -0.09 (15625)+0.1 (625)+248 (25)-600$

Continue the calculation.
$(25) = -1406.25+62.5+6200-600$

Calculate the final value of $(25)$ .
$(25) = \$4256.25$ This is the profit when $x=25$ hundred cell phones are sold.

To interpret $(25)$ , we note that $(25)$ is the profit when $x=25$ hundred cell phones are sold. This means that when2500 cell phones are sold, the profit is \$4256.25.