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Nitella [24]
3 years ago
15

The cost function for a certain company is C = 40x + 800 and the revenue is given by R = 100x − 0.5x2. (THIS IS TO THE SECOND PO

WER) Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $800.

Mathematics
1 answer:
Degger [83]3 years ago
3 0

Answer:

  x = {40, 80}

Step-by-step explanation:

The profit equation is ...

  P = R - C = 100x -0.5x^2 -(40x +800)

  P = -0.5x^2 +60x -800

We want to find values of x such that P = 800

  800 = -0.5x^2 +60x -800

  1600 = -0.5x^2 +60x . . . . add 800

  -3200 = x^2 -120x . . . . . . . multiply by -2

  400 = x^2 -120x +3600 . . . . add 3600 to complete the square

  400 = (x -60)^2 . . . . . . write as a square

  ±20 = x -60 . . . . . . . . . take the square root

  x = 60 ± 20 = {40, 80}

The values of production level (x) that will create profit of $800 are 40 and 80 units.

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Answer:

Over the interval [2, 4], the local minimum is –8.

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Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

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Step-by-step explanation:

If we count 1 to the right on the x-axis and count 3 up on the y-axis, we will end up at the point of intersection which is (1,3).

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