Applications of diffrentiation

Here is a question that I got stuck in..

Can you please help me solve it..

A manager of a supermarket usually adds a mark-up of 20% to the wholesale prices of all the goods he sells.He reckons that he has a loyal core of F customers and that<if he lowers his mark-up to x% he will attract an extra k(20-x) customers from his rivals.Each week the average shopper buys goods whose wholesale value is A pounds.Show that with a mark-up of x% the supermarket will have an anticipated weekly profit of

1/100Ax((F+20k)-kx)) pounds

Show that the manager can increase his profit by reducing his mark-up below 20% provided that 20k>F

Re: Applications of diffrentiation

With a markup of *x*%, the number of customers $\displaystyle C$ is given by:

$\displaystyle C=F+k(20-x)$

With each customer buying an average of ŁA in goods per week, the weekly revenue $\displaystyle R$ in pounds is then:

$\displaystyle R=AC=A(F+k(20-x))$

The weekly profit $\displaystyle P$ is then:

$\displaystyle P=\frac{x}{100}\cdot R=\frac{Ax}{100}(F+k(20-x))=\frac{1}{100}Ax((F+20k)-kx)$

In order to find the markup which yields the maximum profit, I would write the profit function as:

$\displaystyle P(x)=\frac{A}{100}((F+20k)x-kx^2)$

Now, either equate the first derivative to zero to find the critical markup, or find the axis of symmetry for the parabolic profit function.

Re: Applications of diffrentiation

Thanks ^^

After I differentiate it I got this answers, which is the markup that would give us maximum profit

Is that what we need to show??

p(x)=(FAx+20Akx-2Akx^2)/100

dp/dx = (FA+20Ak-2Akx)/100

0=FA+20Ak-2Akx => 0=F+20k-2kx => 20k=2kx-F

Re: Applications of diffrentiation

Once you determine the maximum profit, you can then compare this to the profit when x = 20, and this is one way to show the profit can be increased by reducing the markup. Another, perhaps simpler way is to show that the marginal profit is negative at x = 20. Do you see why this works?

Re: Applications of diffrentiation

Quote:

Originally Posted by

**MarkFL2** Once you determine the maximum profit, you can then compare this to the profit when x = 20, and this is one way to show the profit can be increased by reducing the markup. Another, perhaps simpler way is to show that the marginal profit is negative at x = 20. Do you see why this works?

Umm, Iam kind of confused.

How do we determine the maximum profit?

In terms of k,F and x??

I still don't understand how we can show that It can be increased by reducing the markup :(

Iam sorry,but can you please explain what you mean..

And thanks a lot for your help

Re: Applications of diffrentiation

In order to apply differentiation (as your topic title suggests) you need to equate the first derivative of the profit function to zero to find values of x (critical values) where extrema may occur. Once you have found the critical value(s), you should then employ either the first or second derivative test to determine those that occur at maxima.

Since the profit function is parabolic, and the coefficient of the squared term is negative, how many and what type of extrema should you expect to find?

Re: Applications of diffrentiation

Quote:

Originally Posted by

**MarkFL2** In order to apply differentiation (as your topic title suggests) you need to equate the first derivative of the profit function to zero to find values of x (critical values) where extrema may occur. Once you have found the critical value(s), you should then employ either the first or second derivative test to determine those that occur at maxima.

Since the profit function is parabolic, and the coefficient of the squared term is negative, how many and what type of extrema should you expect to find?

ِMaximum ,, I understand the second way.

Its much easier..

Can you please show me how to find the second derivative..

I know that you need to differentiate this expression (20k=2kx-F) but Iam not sure how you do that

Do you get

2k ??

And how do we know if 2k is max

Re: Applications of diffrentiation

We have the profit function:

$\displaystyle P(x)=\frac{A}{100}((F+20k)x-kx^2)$

If I were going to find the markup which yields the maximum profit, I would proceed as follows:

Equate the first derivative to zero:

$\displaystyle P'(x)=\frac{A}{100}\left((F+20k)-2kx \right)=0$

$\displaystyle x=\frac{F+20k}{2k}$

We can show that this is the maximum profit, since:

$\displaystyle P''(x)=-\frac{Ak}{50}$

and we have $\displaystyle 0<A,k$

So the profit function is concave down for all $\displaystyle x$, showing its extremum as a maximum.

Now, we need to show:

$\displaystyle \frac{F+20k}{2k}<20$

$\displaystyle F+20k<40k$

$\displaystyle F-20k<0$

And since we are provided with the condition $\displaystyle 20k>F$, we have demonstrated that the manager will increase his profit by reducing the markup.

Alternately, we may observe that at a markup of 20%, the marginal profit is:

$\displaystyle P'(20)=\frac{A}{100}\left((F+20k)-2k(20) \right)=\frac{A(F-20k)}{100}$

Now, since we have $\displaystyle 20k>F$, we see the marginal profit is negative at $\displaystyle x=20$, which means the profit is decreasing at that point, which means the manager can increase the profit by lowering the markup.