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2 years ago

For a certain company, the cost for producing x items is 60x+300 and the revenue for selling x items is 100x−0.5x2 .

Mathematics

1 answer:

Katen [24]2 years ago

7 0

The answers are

A. The equation for the profit from selling the items is P = 40x-0.5x²-300.
b. The values of x are 10 and 70.
c. It is not possible to have the profit of 2500

<h3>How to solve for the expression</h3>

We have the cost function in this question to be

C(x) = 60x+300

We have the function of the revenue to be

Revenue = 100x−0.5x²

A. The formula for revenue function is given

revenue - cost

This is expressed as

= (100x−0.5x²)-60x+300

We have to collect like terms and open the equation above.

This given us:

B. when profit = 50$

We have p = 40x-0.5x²-300

50 = 40x-0.5x²-300

Multiply the two sides by 10

This gives

500 = 400x - 5x² - 3000

This gives a quadratic equation

5x² - 400x - 3500 = 0

To solve the equation you have to make use of a quadratic calculator.

This gives us the values

c. We have P = 40x-0.5x²-300.

at P = 2500

2500 = 40x-0.5x²-300.

Multiply the equation by 10

25000 = 400x - 5x² - 3000

collect like terms

400x - 5x2 - 3000 +28000

400x -5x2 +28000

We have to take the discriminant

-400² - 4*5*28000

= -400000

The discriminant is negative hence it is not possible.

Read more on quadratic polynomials here:

brainly.com/question/25841119

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Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .