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KatRina [158]
3 years ago
7

Please help the answer choices are 3 1/103 2/56 5/88 3/4

Mathematics
1 answer:
Lelu [443]3 years ago
8 0
Um hello, ill help you. C:
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5

Step-by-step explanation:

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Number 3 I need some help
Nadusha1986 [10]

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15+10x>_75

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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.

6 0
3 years ago
The graph of f(x)=x2 was transformed to create the graph of g(x)=(x+5)2.
alexandr1967 [171]

Answer:

a horizontal shift left

Step-by-step explanation:

my teacher always tells me whatever you think it is in the parenthisis do the opposite. Sorry not a good explanation

5 0
3 years ago
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