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jarptica [38.1K]
3 years ago
12

A manufacturer finds it costs him x + 5x + 7 dollars to produce x tons of an item. At 2 production levels above 3 tons, he must

hire additional workers, and his costs increase by 3(x - 3) dollars on his total production. If the price he receives is $13 per ton regardless of how much he manufactures and if he has a plant capacity of 10 tons, what level of output maximizes his profits?
Mathematics
1 answer:
garri49 [273]3 years ago
4 0

Answer: The maximum = 3 tons

Step-by-step explanation:

The cost function C(x) = x + 5x + 7

P(×) = 13x - x^2 - 5x -7

If x <3

P= x^2 +8x -7

Differentiating to get x

dp/dx = -2x + 8

X= 8/2

C=4

Maximum will be 3 tons

When x=3

P= 13x - x^2 -5x +7 -3x + 9

When x>3

dp/dx = x^2+5x +2

X = 5/2 = 2.5

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Step-by-step explanation:

1. 5 is the slope because it is the rate. 20 is the y-intercept because that is how much you already have.

2. 20 because that is how much you already have.

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Simplify the expression: –7(5 − u)
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A cylinder has radius r and height h. A. How many times greater is the surface area of a cylinder when both dimensions are multi
Ierofanga [76]

Answer: A. Factor 2 => 4x greater

                   Factor 3 => 9x greater

                   Factor 5 => 25x greater

Step-by-step explanation: A. A cylinder is formed by 2 circles and a rectangle in the middle. That's why surface area is given by circumference of a circle, which is the length of the rectangle times height of the rectangle, i.e.:

A = 2.π.r.h

A cylinder of radius r and height h has area:

A_{1} = 2πrh

If multiply both dimensions <u>by a factor of 2</u>:

A_{2} = 2.π.2r.2h

A_{2} = 8πrh

Comparing A_{1} to A_{2} :

\frac{A_{2}}{A_{1}} = \frac{8.\pi.rh}{2.\pi.rh} = 4

Doubling radius and height creates a surface area of a cylinder 4 times greater.

<u>By factor 3:</u>

A_{3} = 2.\pi.3r.3h

A_{3} = 18.\pi.r.h

Comparing areas:

\frac{A_{3}}{A_{1}} = \frac{18.\pi.r.h}{2.\pi.r.h} = 9

Multiplying by 3, gives an area 9 times bigger.

<u>By factor 5</u>:

A_{5} = 2.\pi.5r.5h

A_{5} = 50.\pi.r.h

Comparing:

\frac{A_{5}}{A_{1}} = \frac{50.\pi.r.h}{2.\pi.r.h} = 25

The new area is 25 times greater.

B. By analysing how many times greater and the factor that the dimensions are multiplied, you can notice the increase in area is factor². For example, when multiplied by a factor of 2, the new area is 4 times greater.

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