Company Mathematics

[mathdad]

The price p (in dollars) and the quantity x sold of a certain product obey the demand equation x = -20p + 500, where 0 < p less than or equal to 25.

A. Find a model that represents the revenue R as a function of x.

Revenue = price x quantity sold or in short.

R = px

B. What price should the company charge to maximize revenue?

Here we go.

Part A

I know that x is given to be
-20p + 500.

R = xp

R = (-20p + 500)p

R = -20p^2 + 500p

Part B

Let p = price company should charge to maximize revenue.

p = -b/2a

p = (-500)/2(-20)

p = -500/-40

p = 500/40

p = 12.50

I know that 0 < 12.50 less than or equal to 25. So, I say my answer is right. What do you say?

 

[MarkFL]

In part A, you've given revenue as a function of price \(p\), when the instructions say to give revenue as a function of \(x\).

 

[mathdad]

In part A, you've given revenue as a function of price \(p\), when the instructions say to give revenue as a function of \(x\).

Corrections will be made and posted. This is common when not paying attention to detail.

 

[mathdad]

In part A, you've given revenue as a function of price \(p\), when the instructions say to give revenue as a function of \(x\).

The price p (in dollars) and the quantity x sold of a certain product obey the demand equation x = -20p + 500, where 0 < p less than or equal to 25.

A. Find a model that represents the revenue R as a function of x.

Revenue = price x quantity sold or in short.

R = px

B. What price should the company charge to maximize revenue?

Here we go.

Part A

I know that x is given to be
-20p + 500.

R = xp

x = -20p + 500

Solving for p now.

x - 500 = -20p

(x - 500)/(-20) = p

R = px

R = [(x - 500)/(-20)]x

R = -(x/20) + 25)x

R = -(x^2)/20 + 25x

Part B

Let x = price company should charge to maximize revenue.

x = -b/2a

x = -(25)/(-1/20)

x = 500

The company should charge 500 dollars to maximize revenue.

 

[MarkFL]

When you found the axis of symmetry, your value of \(x\) was twice the actual value because you dropped the factor of 2 from the denominator. But recall also, that \(x\) is not the price, but rather we find revenue is maximized when \(x=250\) and at that level of output, we find:

[MATH]p=\frac{500-250}{20}=\frac{25}{2}=12.5[/MATH]
You had the correct price in your first attempt, because you actually had revenue as a function of price, but that wasn't what the instructions asked for.

 

[mathdad]

When you found the axis of symmetry, your value of \(x\) was twice the actual value because you dropped the factor of 2 from the denominator. But recall also, that \(x\) is not the price, but rather we find revenue is maximized when \(x=250\) and at that level of output, we find:

[MATH]p=\frac{500-250}{20}=\frac{25}{2}=12.5[/MATH]
You had the correct price in your first attempt, because you actually had revenue as a function of price, but that wasn't what the instructions asked for.

Ok. I guess silly mistakes lead to the wrong answer. Boy, I hate when that happens! You know, I am posting too many questions forcing me to rush through my computation. Less questions posted from now on.

 

[MarkFL]

Don't be too hard on yourself for mistakes...we all make them and while they can be embarrassing (I know I have kicked myself many times for mistakes I've made with problems on the forums) in a public setting, just learn from them and apply the lessons learned to future problems.

 

[Steven G]

The price p (in dollars) and the quantity x sold of a certain product obey the demand equation x = -20p + 500, where 0 < p less than or equal to 25.

A. Find a model that represents the revenue R as a function of x.

Revenue = price x quantity sold or in short.

R = px

B. What price should the company charge to maximize revenue?

Here we go.

Part A

I know that x is given to be
-20p + 500.

R = xp

x = -20p + 500

Solving for p now.

x - 500 = -20p

(x - 500)/(-20) = p

R = px

R = [(x - 500)/(-20)]x

R = -(x/20) + 25)x

R = -(x^2)/20 + 25x

Part B

Let x = price company should charge to maximize revenue.

x = -b/2a

x = -(25)/(-1/20)

x = 500

The company should charge 500 dollars to maximize revenue.

In solving for p in x - 500 = -20p. The lhs has 0p and the rhs has -20p. So the lhs has more p's so bring all p's to the lhs and everything else to the rhs. This way you will end up with 50p on the lhs and when you divide by 50, to get p alone, you will be dividing by a positive number. This will make things easier.

 

[mathdad]

Thank you everyone for your help and guidance.