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harkovskaia [24]
3 years ago
6

The total cost (in dollars) for a company to manufacture and sell x items per week is C = 40 x + 640 , whereas the revenue broug

ht in by selling all x items is R = 66 x − 0.2 x 2 . How many items must be sold to obtain a weekly profit of $ 200 ?
Mathematics
1 answer:
ikadub [295]3 years ago
3 0

Answer:

The company must sell 60 or 70 items to obtain a weekly profit of 200.

Step-by-step explanation:

The profit is the difference between the revenue and the cost of a given task, therefore:

\text{profit} = R - C\\\text{profit} = 66*x - 0.2*x^2 - (40*x + 640)\\\text{profit} = 66*x - 0.2*x^2 - 40*x - 640\\\text{profit} = - 0.2*x^2 + 26*x - 640

To have a profit of 200, we need to sell:

-0.2*x^2 + 26*x - 640 = 200\\-0.2*x^2 + 26*x -840 = 0\text{ } *\frac{-1}{0.2}\\x^2 -130 + -4200 = 0\\x_{1,2} = \frac{-(-130) \pm \sqrt{(-130)^2 - 4*1*(-4200)}}{2*1}\\x_{1,2} = \frac{130 \pm \sqrt{16900 + 16800}}{2}\\x_{1,2} = \frac{130 \pm \sqrt{100}}{2}\\x_{1,2} = \frac{130 \pm 10}{2}\\x_{1} = \frac{130 + 10}{2} = \frac{140}{2} = 70\\ x_{2} = \frac{130 - 10}{2} = \frac{120}{2}  = 60

The company must sell 60 or 70 items to obtain a weekly profit of 200.

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a rectangular storage container without a lid is to have a volume of 10 m3. the length of its base is twice the width. material
victus00 [196]

245.31 (dollars) is the cost (in dollars) of materials for the least expensive such container.

<h3>What is maxima and minima ? </h3>

Calculus maxima and minima are found  using the concept of derivatives. Knowing that the derivative concept gives information about the slope/slope of a function, we find the point where the slope is zero. These points are called inflection points/stationary points. These are the points associated with the maximum or minimum (local) values ​​of the function.

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<h3>Calculation</h3>

Suppose the width is x (m), length of the base is 2x (m), the base area is 2x^2 (m^2).

Since the volume is 10 (m^3), the height has to be 10/2x^2 (m) = 5/x^2.

The cost of making such container is

cost of base: 2x^2*15 = 30x^2

cost of sides: (2*2x*5/x^2 + 2*x*5/x^2)*9 = 270/x

The overall cost is hence the sum of the base and the sides: f(x) = 30x^2 + 270/x

The get the minimum,

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f(x) = 245.31 (dollars)

learn more about maxima and minima here :

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