What is the optimal price for maximum profit?

[Karol]

Homework Statement

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling

The Attempt at a Solution

The profit S:
$$S=nx=\left[ \frac{a}{x-c}+\frac{b}{100-x} \right]x$$
$$S'=\frac{a}{x-c}+\frac{b}{100-x}+\left[ \frac{-a}{(x-c)^2}+\frac{-b}{(100-x)^2} \right]x$$
$$\rightarrow \frac{a}{x-c}+\frac{b}{100-x}=\left[ \frac{-a}{(x-c)^2}+\frac{-b}{(100-x)^2} \right]x$$

 

[Buzz Bloom]

Hi Karol:

I see two issues with your solution attempt.
1. I suggest you write the S equation as

S = x [a/(x-c) - b(x-100)]​

2. You want to solve the following equation for x:

S' = 0.​

The final equation you have does not show you have done that.

Hope this h elps.

Regards,
Buzz

 

[DoItForYourself]

In order to get the profit function, you must subtract the total manufacturing and distribution cost from the money you get by selling them.

When you form the function, then you set the first derivative of the function to be equal to zero. But, be careful, that does not mean that your profits will be at their maximum value. They can also be at their minimum value.

 

[Ray Vickson]

Homework Statement

View attachment 214814

Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling

The Attempt at a Solution

The profit S:
$$S=nx=\left[ \frac{a}{x-c}+\frac{b}{100-x} \right]x$$
$$S'=\frac{a}{x-c}+\frac{b}{100-x}+\left[ \frac{-a}{(x-c)^2}+\frac{-b}{(100-x)^2} \right]x$$
$$\rightarrow \frac{a}{x-c}+\frac{b}{100-x}=\left[ \frac{-a}{(x-c)^2}+\frac{-b}{(100-x)^2} \right]x$$

(1) The question said that
$$n = \frac{a}{x-c} + b(100-x),$$
which is not what you wrote.

(2) Profit = revenue - cost; you are attempting to maximize revenue, not profit.

 

[Karol]

$$n = \frac{a}{x-c} + b(100-x)$$
$$Profit=n(x-c)=\left[ \frac{a}{x-c} + b(100-x) \right](x-c)=a+b(100-x)(x-c)$$
$$P'=b[-x(x-c)+100-x]=b[-x^2+x(c-1)+100]$$
I can't find the roots for P'=0

 

[Ray Vickson]

$$n = \frac{a}{x-c} + b(100-x)$$
$$Profit=n(x-c)=\left[ \frac{a}{x-c} + b(100-x) \right](x-c)=a+b(100-x)(x-c)$$
$$P'=b[-x(x-c)+100-x]=b[-x^2+x(c-1)+100]$$
I can't find the roots for P'=0

Easy: just solve a quadratic equation---done back in Algebra I.

However, you made an error: your profit function is quadratic in $x$, so its first derivative is linear in $x.$ You just need to solve a linear equation.

 

[Karol]

$$P=a+b(100-x)(x-c)~\rightarrow P'=b(-2x+100+c)$$
$$P'=0~\rightarrow x=\frac{100+c}{2}$$
Thank you Buzz, Dolt and Ray

 

[DoItForYourself]

You are welcome.

Just do not forget to show that P'(x)>0 for x < $\frac {100+c} {2}$ and P'(x)<0 for x > $\frac {100+c} {2}$.

This is the complete way to show that P(x) is maximized and not minimized for this x.