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Juliette [100K]
3 years ago
14

please help! at a recent cross country meet, spectators had to pay to get in. it cost $7 for adults and $3 for students. the boo

ster club selling tickets let's forgot to keep track of how many adults and student tickets they sold. however, they did know they sold 155 total tickets and made $645. write a system of equations to model this situation
Mathematics
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Substitute 1 in for m.

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QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu
Harman [31]

Answer:

(a)Revenue function, R(x)=580x-x^2

Marginal Revenue function, R'(x)=580-2x

(b)Fixed cost =900 .

Marginal Cost Function=300+50x

(c)Profit,P(x)=-35x^2+280x-900

(d)x=4

Step-by-step explanation:

<u>Part A </u>

Price Function= 580 - 10x

The revenue function

R(x)=x\cdot (580-10x)\\R(x)=580x-x^2

The marginal revenue function

\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x

<u>Part B </u>

<u>(Fixed Cost)</u>

The total cost function of the company is given by c=(30+5x)^2

We expand the expression

(30+5x)^2=(30+5x)(30+5x)=900+300x+25x^2

Therefore, the fixed cost is 900 .

<u> Marginal Cost Function</u>

If  c=900+300x+25x^2

Marginal Cost Function, \frac{dc}{dx}= (900+300x+25x^2)'=300+50x

<u>Part C </u>

<u>Profit Function </u>

Profit=Revenue -Total cost

580x-10x^2-(900+300x+25x^2)\\580x-10x^2-900-300x-25x^2\\$Profit,P(x)=-35x^2+280x-900

<u> Part D </u>

To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.

P(x)=-35x^2+280x-900\\P'(x)=-70x+280\\-70x+280=0\\-70x=-280\\$Divide both sides by -70\\x=4

The number of cakes that maximizes profit is 4.

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