QuestionFind the weekly profit function $P(x)$ for a sandwich store given cost $C(x)=555.00+0.45x$ and revenue $R(x)=-0.001x^{2}+3x$ . Simplify your answer.

Studdy Solution

STEP 1

Assumptions1. The cost function is $C(x)=555.00+0.45 x$ . The revenue function is $R(x)=-0.001 x^{}+3 x$
3. The profit function is the difference between the revenue and cost functions, i.e., $(x) = R(x) - C(x)$

STEP 2

We start by writing down the profit function $(x)$ as the difference between the revenue function $R(x)$ and the cost function $C(x)$ .
$(x) = R(x) - C(x)$

STEP 3

Now, we substitute the given functions $R(x)$ and $C(x)$ into the profit function $(x)$ .
$(x) = (-0.001 x^{2}+3 x) - (555.00+0.45 x)$

STEP 4

Next, we distribute the negative sign on the right side of the equation to each term inside the parentheses.
$(x) = -0.001 x^{2}+3 x -555.00 -0.45 x$

STEP 5

Now, we combine like terms in the equation.
$(x) = -0.001 x^{2} + (3 -0.45) x -555.00$

STEP 6

Finally, we simplify the equation to get the profit function $(x)$ .
$(x) = -0.001 x^{2} +2.55 x -555.00$ So, the store's weekly profit function,, is $(x) = -0.001 x^{2} +2.55 x -555.00$ .

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QuestionFind the weekly profit function $P(x)$ for a sandwich store given cost $C(x)=555.00+0.45x$ and revenue $R(x)=-0.001x^{2}+3x$ . Simplify your answer.

Studdy Solution

STEP 1

Assumptions1. The cost function is $C(x)=555.00+0.45 x$ . The revenue function is $R(x)=-0.001 x^{}+3 x$
3. The profit function is the difference between the revenue and cost functions, i.e., $(x) = R(x) - C(x)$

STEP 2

We start by writing down the profit function $(x)$ as the difference between the revenue function $R(x)$ and the cost function $C(x)$ .
$(x) = R(x) - C(x)$

STEP 3

Now, we substitute the given functions $R(x)$ and $C(x)$ into the profit function $(x)$ .
$(x) = (-0.001 x^{2}+3 x) - (555.00+0.45 x)$

STEP 4

Next, we distribute the negative sign on the right side of the equation to each term inside the parentheses.
$(x) = -0.001 x^{2}+3 x -555.00 -0.45 x$

STEP 5

Now, we combine like terms in the equation.
$(x) = -0.001 x^{2} + (3 -0.45) x -555.00$

STEP 6

Finally, we simplify the equation to get the profit function $(x)$ .
$(x) = -0.001 x^{2} +2.55 x -555.00$ So, the store's weekly profit function,, is $(x) = -0.001 x^{2} +2.55 x -555.00$ .

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QuestionFind the weekly profit function P(x)P(x)P(x) for a sandwich store given cost C(x)=555.00+0.45xC(x)=555.00+0.45xC(x)=555.00+0.45x and revenue R(x)=−0.001x2+3xR(x)=-0.001x^{2}+3xR(x)=−0.001x2+3x. Simplify your answer.

Studdy Solution

STEP 1

STEP 2

We start by writing down the profit function (x)(x)(x) as the difference between the revenue function R(x)R(x)R(x) and the cost function C(x)C(x)C(x).(x)=R(x)−C(x)(x) = R(x) - C(x)(x)=R(x)−C(x)

STEP 3

Now, we substitute the given functions R(x)R(x)R(x) and C(x)C(x)C(x) into the profit function (x)(x)(x).(x)=(−0.001x2+3x)−(555.00+0.45x)(x) = (-0.001 x^{2}+3 x) - (555.00+0.45 x)(x)=(−0.001x2+3x)−(555.00+0.45x)

STEP 4

Next, we distribute the negative sign on the right side of the equation to each term inside the parentheses.(x)=−0.001x2+3x−555.00−0.45x(x) = -0.001 x^{2}+3 x -555.00 -0.45 x(x)=−0.001x2+3x−555.00−0.45x

STEP 5

Now, we combine like terms in the equation.(x)=−0.001x2+(3−0.45)x−555.00(x) = -0.001 x^{2} + (3 -0.45) x -555.00(x)=−0.001x2+(3−0.45)x−555.00

STEP 6

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QuestionFind the weekly profit function $P(x)$ for a sandwich store given cost $C(x)=555.00+0.45x$ and revenue $R(x)=-0.001x^{2}+3x$ . Simplify your answer.

We start by writing down the profit function $(x)$ as the difference between the revenue function $R(x)$ and the cost function $C(x)$ .
$(x) = R(x) - C(x)$

Now, we substitute the given functions $R(x)$ and $C(x)$ into the profit function $(x)$ .
$(x) = (-0.001 x^{2}+3 x) - (555.00+0.45 x)$

Next, we distribute the negative sign on the right side of the equation to each term inside the parentheses.
$(x) = -0.001 x^{2}+3 x -555.00 -0.45 x$

Now, we combine like terms in the equation.
$(x) = -0.001 x^{2} + (3 -0.45) x -555.00$