Business Math Algebra

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mathdad

mathdad

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The price p (in dollars) and the quantity x sold of a certain product obey the demand equation x = -5p + 100, where 0 < p ≤ 20.

Note: Revenue = xp

See my work per part.

A. Express the revenue R as a function of x. [R(x)].

Solution:

R(x) = xp

Solve the demand equation for p.

x = -5p + 100

x - 100 = -5p

(x - 100)/-5 = p

(-1/5)(x - 100) = p

(-x/5) + 20 = p

R(x) = xp

R(x) = x[(-x/5) + 20]

R(x) = (-x^2)/5 + 20x

B. What is the revenue R if 15 units are sold?

Solution:

I need to find R(15).

R(15) = -(15)^2 + 20(15)

R(15) = 75 dollars

C. What price p (in dollars) should the company charge to maximize revenue?

I know that x = -b/2a is used here, right? Should I be looking for p = -b/2a?

D. What price p (in dollars) should the company charge to earn at least $480 in revenue?

What is the set up for part D?
 
mathdad said:
The price p (in dollars) and the quantity x sold of a certain product obey the demand equation x = -5p + 100, where 0 < p ≤ 20.

Note: Revenue = xp

See my work per part.

A. Express the revenue R as a function of x. [R(x)].

Solution:

R(x) = xp

Solve the demand equation for p.

x = -5p + 100

x - 100 = -5p

(x - 100)/-5 = p

(-1/5)(x - 100) = p

(-x/5) + 20 = p

R(x) = xp

R(x) = x[(-x/5) + 20]

R(x) = (-x^2)/5 + 20x

B. What is the revenue R if 15 units are sold?

Solution:

I need to find R(15).

R(15) = -(15)^2 + 20(15).....................................................................................Incorrect

R(15) = 75 dollars

C. What price p (in dollars) should the company charge to maximize revenue?

I know that x = -b/2a is used here, right? Should I be looking for p = -b/2a?

D. What price p (in dollars) should the company charge to earn at least $480 in revenue?

What is the set up for part D?
Click to expand...
R(x) = (-x^2)/5 + 20x

R(15) = (-15^2)/5 + 20*15 = -45 +300 =255

D. What is the amount sold(x, x>0) for R= $480 ?
 
I hate these questions that ask for optimization of a quadratic without calculus. Yes, there is a solution available without calculus: x has its finite extremum at x = - b/2a. But why teach a method for finding a finite extremum for an extremely narrow class of functions? No one who does not know differential calculus or integer programming is likely to be asked to cope with such problems.
 
mathdad said:
... D. What price p (in dollars) should the company charge to earn at least $480 in revenue?

What is the set up ...?
Click to expand...
What is the expression for revenue?

What does it look like, to set an expression to be "at least 480"?

?
 
C) Don't stress out yet that you are being asked for p. You want the max Revenue. You have R in terms of x. You can find the x value that gives the max R using your formula x=-b/2a. Once you have that x value you can turn it into a p value by using p=(-x/5) + 20

I told you a few times now that it really really is better to bring the variable you are trying to solve for (in this case it was p) to the side which has more p's. This way you will NEVER end up with a negative number of p's.

In the equation x = -5p + 100, the lhs has 0 p's while the rhs side has -5 p's. Since 0>-5, you bring the p's to the lhs. This will give you 5p = 100 - x or p = 20 - x/5
 
Last edited:
JeffM said:
I hate these questions that ask for optimization of a quadratic without calculus. Yes, there is a solution available without calculus: x has its finite extremum at x = - b/2a. But why teach a method for finding a finite extremum for an extremely narrow class of functions? No one who does not know differential calculus or integer programming is likely to be asked to cope with such problems.
Click to expand...

I know the basics of differential calculus. How is calculus used to answer this question? What are the steps?
 
Jomo said:
C) Don't stress out yet that you are being asked for p. You want the max Revenue. You have R in terms of x. You can find the x value that gives the max R using your formula x=-b/2a. Once you have that x value you can turn it into a p value by using p=(-x/5) + 20

I told you a few times now that it really really is better to bring the variable you are trying to solve for (in this case it was p) to the side which has more p's. This way you will NEVER end up with a negative number of p's.

In the equation x = -5p + 100, the lhs has 0 p's while the rhs side has -5 p's. Since 0>-5, you bring the p's to the lhs. This will give you 5p = 100 - x or p = 20 - x/5
Click to expand...
Let x = -b/2a

x = -20/2(-1/5)

x = -20/(-2/5)

x = 20/(2/5)

x = 100/2

x = 50

Now use 5p = 100 - x to solve for p (in dollars).

5p = 100 - 50

5p = 50

p = 50/5

p = 10 dollars
 
mathdad said:
I know the basics of differential calculus. How is calculus used to answer this question? What are the steps?
Click to expand...
Probably 95% or more of the practical use of differential calculus is to find extrema, meaning minima or minima.

Assuming f(x) is differentiable on (a, b), f'(x) is the first derivative, and f''(x) is the second derivative, f'(x) will = 0 at an extremum in the interval, f''(x) will not be less than 0 at a minimum, and f''(x) will not be greater than 0 at a maximum. So you need remember nothing about parabolas to deal with quadratics. You just remember the few simple rules to compute derivative that apply to all differentiable functions. There are particularly easy for quadratics.

f(x) = ax^2 + bx + c leads to f'(x) = 2ax + b, which equals 0 iff x = -b/2a, and to f''(x) = 2a, which is not zero if f(x) is not linear.

Virtually all of basic economics is based on differential calculus.
 
See my work per part.

A. Express the revenue R as a function of x. [R(x)].

Solution:

R(x) = xp

Solve the demand equation for p.

x = -5p + 100

x - 100 = -5p

(x - 100)/-5 = p

(-1/5)(x - 100) = p

(-x/5) + 20 = p

R(x) = xp

R(x) = x[(-x/5) + 20]

R(x) = (-x^2)/5 + 20x


B. What is the revenue R if 15 units are sold?

Solution:

I need to find R(15).

R(15) = -(15)^2/5 + 20(15)7

R(15) = 255 dollars

C. What price p (in dollars) should the company charge to maximize revenue?

Let x = -b/2a

x = -20/2(-1/5)

x = -20/(-2/5)

x = 20/(2/5)

x = 100/2

x = 50

Now use 5p = 100 - x.

5p = 100 - 50

5p = 50

p = 50/5

p = 10 dollars

D. What price p (in dollars) should the company charge to earn at least $480 in revenue?

x = -5p + 100

R(p) = (-5p + 100)p

R(p) = -5p^2 + 100p

Let R(p) = 480

480 = -5p^2 + 100p

5p^2 - 100p + 480 = 0

I got p = 8 and p = 12.
 
JeffM said:
Probably 95% or more of the practical use of differential calculus is to find extrema, meaning minima or minima.

Assuming f(x) is differentiable on (a, b), f'(x) is the first derivative, and f''(x) is the second derivative, f'(x) will = 0 at an extremum in the interval, f''(x) will not be less than 0 at a minimum, and f''(x) will not be greater than 0 at a maximum. So you need remember nothing about parabolas to deal with quadratics. You just remember the few simple rules to compute derivative that apply to all differentiable functions. There are particularly easy for quadratics.

f(x) = ax^2 + bx + c leads to f'(x) = 2ax + b, which equals 0 iff x = -b/2a, and to f''(x) = 2a, which is not zero if f(x) is not linear.

Virtually all of basic economics is based on differential calculus.
Click to expand...

Good notes for my future study of differential calculus. However, I am currently reviewing college algebra essentials. I just posted my solution to parts A through D. Am I correct per part?
 
mathdad said:
What is left to do with part D?
Click to expand...
Part D asks what price the company needs for earning at least $480 in revenue.

The phrase 'at least' means any revenue amount $480 or higher.

All of the values from p=8 through p=12 cause revenue to be $480 or higher.


mathdad said:
... I got p = 8 and p = 12.
Click to expand...
You reported only the values of p that cause revenue to be exactly $480.

I typed the correct answer in post #13. Did you notice it?

You can see how R(p) is 480 or higher in the interval [8,12] by looking at a graph. Google the following plot command.

plot y=-5*x^2 + 100*x, y=480

?
 
Otis said:
Part D asks what price the company needs for earning at least $480 in revenue.

The phrase 'at least' means any revenue amount $480 or higher.

All of the values from p=8 through p=12 cause revenue to be $480 or higher.



You reported only the values of p that cause revenue to be exactly $480.

I typed the correct answer in post #13. Did you notice it?

You can see how R(p) is 480 or higher in the interval [8,12] by looking at a graph. Google the following plot command.

plot y=-5*x^2 + 100*x, y=480

?
Click to expand...

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mathdad said:
I am not doing anymore math until next week. I want to keep my word.
Click to expand...
I do not understand your response to my question.

Please answer 'yes' or 'no'. Are you finished with this exercise?

\(\;\)
 
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