Questiona. Write the cost function $C(x)=565.00+0.65 x$ for $x$ roast beef sandwiches. b. Profit function is $P(x)=-0.001 x^{2}+2.35 x-565.00$ . c. Find $x$ for max profit and state max profit $\$ \square$ when $\square$ sandwiches sold.

Studdy Solution

STEP 1

Assumptions1. The cost function is given by $C(x)=565.00+0.65x$ , where $x$ is the number of roast beef sandwiches made and sold each week. . The revenue function is given by $R(x)=-0.001x^{}+3x$ .
3. The profit function is the difference between revenue and cost, given by $(x)=R(x)-C(x)$ .

STEP 2

We are given the cost function $C(x)$ and the revenue function $R(x)$ . We need to find the profit function $(x)$ , which is the difference between the revenue and the cost.
$(x) = R(x) - C(x)$

STEP 3

Substitute the given functions $R(x)$ and $C(x)$ into the equation for $(x)$ .
$(x) = (-0.001x^{2}+3x) - (565.00+0.65x)$

STEP 4

implify the equation by combining like terms.
$(x) = -0.001x^{2} +3x -0.65x -565.00$

STEP 5

Further simplify the equation.
$(x) = -0.001x^{2} +2.35x -565.00$

STEP 6

To find the number of roast beef sandwiches that should be sold to maximize profit, we need to find the derivative of the profit function and set it equal to zero. This is because the maximum or minimum of a function occurs where its derivative is zero.
$'(x) = -0.002x +2.35$

STEP 7

Set the derivative equal to zero and solve for $x$ .
$-0.002x +2.35 =0$

STEP 8

olve for $x$ .
$x = \frac{2.35}{0.002}$

STEP 9

Calculate the value of $x$ .
$x =1175$

STEP 10

Substitute $x =1175$ into the profit function $(x)$ to find the maximum weekly profit.
$(1175) = -0.001(1175)^{2} +2.35(1175) -565.00$

STEP 11

Calculate the value of $(1175)$ .
$(1175) = \$103.25$ The maximum weekly profit is $103.25 when1175 sandwiches are sold.

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Questiona. Write the cost function $C(x)=565.00+0.65 x$ for $x$ roast beef sandwiches. b. Profit function is $P(x)=-0.001 x^{2}+2.35 x-565.00$ . c. Find $x$ for max profit and state max profit $\$ \square$ when $\square$ sandwiches sold.

Studdy Solution

STEP 1

Assumptions1. The cost function is given by $C(x)=565.00+0.65x$ , where $x$ is the number of roast beef sandwiches made and sold each week. . The revenue function is given by $R(x)=-0.001x^{}+3x$ .
3. The profit function is the difference between revenue and cost, given by $(x)=R(x)-C(x)$ .

STEP 2

We are given the cost function $C(x)$ and the revenue function $R(x)$ . We need to find the profit function $(x)$ , which is the difference between the revenue and the cost.
$(x) = R(x) - C(x)$

STEP 3

Substitute the given functions $R(x)$ and $C(x)$ into the equation for $(x)$ .
$(x) = (-0.001x^{2}+3x) - (565.00+0.65x)$

STEP 4

implify the equation by combining like terms.
$(x) = -0.001x^{2} +3x -0.65x -565.00$

STEP 5

Further simplify the equation.
$(x) = -0.001x^{2} +2.35x -565.00$

STEP 6

To find the number of roast beef sandwiches that should be sold to maximize profit, we need to find the derivative of the profit function and set it equal to zero. This is because the maximum or minimum of a function occurs where its derivative is zero.
$'(x) = -0.002x +2.35$

STEP 7

Set the derivative equal to zero and solve for $x$ .
$-0.002x +2.35 =0$

STEP 8

olve for $x$ .
$x = \frac{2.35}{0.002}$

STEP 9

Calculate the value of $x$ .
$x =1175$

STEP 10

Substitute $x =1175$ into the profit function $(x)$ to find the maximum weekly profit.
$(1175) = -0.001(1175)^{2} +2.35(1175) -565.00$

STEP 11

Calculate the value of $(1175)$ .
$(1175) = \$103.25$ The maximum weekly profit is $103.25 when1175 sandwiches are sold.

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Studdy Solution

STEP 1

STEP 2

We are given the cost function C(x)C(x)C(x) and the revenue function R(x)R(x)R(x). We need to find the profit function (x)(x)(x), which is the difference between the revenue and the cost.(x)=R(x)−C(x)(x) = R(x) - C(x)(x)=R(x)−C(x)

STEP 3

Substitute the given functions R(x)R(x)R(x) and C(x)C(x)C(x) into the equation for (x)(x)(x).(x)=(−0.001x2+3x)−(565.00+0.65x)(x) = (-0.001x^{2}+3x) - (565.00+0.65x)(x)=(−0.001x2+3x)−(565.00+0.65x)

STEP 4

implify the equation by combining like terms.(x)=−0.001x2+3x−0.65x−565.00(x) = -0.001x^{2} +3x -0.65x -565.00(x)=−0.001x2+3x−0.65x−565.00

STEP 5

Further simplify the equation.(x)=−0.001x2+2.35x−565.00(x) = -0.001x^{2} +2.35x -565.00(x)=−0.001x2+2.35x−565.00

STEP 6

STEP 7

Set the derivative equal to zero and solve for xxx.−0.002x+2.35=0-0.002x +2.35 =0−0.002x+2.35=0

STEP 8

olve for xxx.x=2.350.002x = \frac{2.35}{0.002}x=0.0022.35​

STEP 9

Calculate the value of xxx.x=1175x =1175x=1175

STEP 10

Substitute x=1175x =1175x=1175 into the profit function (x)(x)(x) to find the maximum weekly profit.(1175)=−0.001(1175)2+2.35(1175)−565.00(1175) = -0.001(1175)^{2} +2.35(1175) -565.00(1175)=−0.001(1175)2+2.35(1175)−565.00

STEP 11

Calculate the value of (1175)(1175)(1175).(1175)=$103.25(1175) = \$103.25(1175)=$103.25The maximum weekly profit is $103.25 when1175 sandwiches are sold.

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We are given the cost function $C(x)$ and the revenue function $R(x)$ . We need to find the profit function $(x)$ , which is the difference between the revenue and the cost.
$(x) = R(x) - C(x)$

Substitute the given functions $R(x)$ and $C(x)$ into the equation for $(x)$ .
$(x) = (-0.001x^{2}+3x) - (565.00+0.65x)$

implify the equation by combining like terms.
$(x) = -0.001x^{2} +3x -0.65x -565.00$

Further simplify the equation.
$(x) = -0.001x^{2} +2.35x -565.00$

Set the derivative equal to zero and solve for $x$ .
$-0.002x +2.35 =0$

olve for $x$ .
$x = \frac{2.35}{0.002}$

Calculate the value of $x$ .
$x =1175$

Substitute $x =1175$ into the profit function $(x)$ to find the maximum weekly profit.
$(1175) = -0.001(1175)^{2} +2.35(1175) -565.00$

Calculate the value of $(1175)$ .
$(1175) = \$103.25$ The maximum weekly profit is $103.25 when1175 sandwiches are sold.