• Jul 27th 2006, 11:45 PM
A restaurant chain plans to introduce a new buffalo steak dinner. Managers tested various prices in their establishments and arrived at the following estimates:

______________________________________________
Price "p" \$14.95 \$19.95 \$24.95 \$29.95
______________________________________________
Number sold
each week "x" 2800 2300 1600 300
______________________________________________

a) Fit a quadratic function that gives demand as a function of price.

b) Find the price per dinner that results in the sale of 2600 buffalo steak dinners.

c) Using the function found in part (a), find a function for revenue as a function of price.

d) Plot the graph of revenue.

e) Using derivatives, find the price at which revenue is maximized. What is the maximum revenue? How many buffalo dinners will be sold each week at this price?

f) If menus are printed with the optimal price (found in part (e)) mistakenly reduced by \$1, what is the loss in revenue? Is this a big deal? Should you fire the printer?
• Jul 28th 2006, 07:25 AM
ThePerfectHacker
Using my software, I found that
$y=-1.83x^2-.29x+30.18$ is and excellent match.
• Jul 28th 2006, 08:51 AM
Quote:

Originally Posted by ThePerfectHacker
Using my software, I found that
$y=-1.83x^2-.29x+30.18$ is and excellent match.

And there's absolutely no other way to do this problem??

p.s. What is this software of yours anyway??? LOL
• Jul 28th 2006, 09:35 AM
ThePerfectHacker
Quote:

And there's absolutely no other way to do this problem??

p.s. What is this software of yours anyway??? LOL

Of course there is, it is just extremely messy which is why I used software.

• Jul 28th 2006, 10:47 AM
Quote:

Originally Posted by ThePerfectHacker
Of course there is, it is just extremely messy which is why I used software.

I still dont know how you managed to solve that problem using what you showed me :-(
• Jul 28th 2006, 11:42 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Using my software, I found that
$y=-1.83x^2-.29x+30.18$ is and excellent match.

This is not a good predictor of sales from price.

Code:

```>x=[14.95;19.95;24.95;29.95]         14.95         19.95         24.95         29.95 >y=-1.838*x^2-0.29*x+30.18     -384.953     -707.134     -1121.22       -1627.2 > > >x|y         14.95      -384.953         19.95      -707.134         24.95      -1121.22         29.95      -1627.2 >```
A better model can be obtained using the Excel Solver, this gives
a solution:

$
sales=1650.22+195.16\ price - 7.999\ price^2
$

This gives:

Code:

```> >price=[14.95;19.95;24.95;29.95]         14.95         19.95         24.95         29.95 >sales=-7.999*price^2+195.16*price+1650.22       2780.07       2360.04       1540.06       320.139 > > >price|sales         14.95      2780.07         19.95      2360.04         24.95      1540.06         29.95      320.139 >```
A plot comparing this model with the estimated sales is shown in the attachment

RonL
• Jul 28th 2006, 12:35 PM
ThePerfectHacker
The quadradic which I posted is the parabola of best fit. How can you have a better one?
• Jul 28th 2006, 01:28 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
The quadradic which I posted is the parabola of best fit. How can you have a better one?

Because it is not a good fit to what was required. Without checking or any
real knowlege I would bet use have used sales as the independent variable
rather than price.

RonL
• Jul 29th 2006, 12:25 AM
So Excel solver and Graphing software.

Can someone please explain to me how I wld ago about doing this problem if I was attempting to do it without using either of the tools.

Thank you
• Jul 29th 2006, 01:04 AM
CaptainBlack
Quote:

So Excel solver and Graphing software.

Can someone please explain to me how I wld ago about doing this problem if I was attempting to do it without using either of the tools.

Thank you

With the Excel solver we are solving a non-linear regression problem, which
for practical purposes is impossible with pencil and paper.

There are method which use linear regression to fit a polynomial, but these
are not widely publicised these days as far as I can see.

It can be done for this data by trial and error without too much trouble, or
it can be done with orthogonal polynomials.

Other than trial and error these can be searched for on Wikipedia, but I
think you will not find a lot of practical help there.

RonL
• Jul 29th 2006, 08:26 AM
Excel Solver
Ok.

Er, how would you use the Excel Solver to solve this problem then?

Sorry but I really dont know how to use Excel to solve these problems.
• Jul 29th 2006, 09:13 AM
CaptainBlack
Quote:

Ok.

Er, how would you use the Excel Solver to solve this problem then?

Sorry but I really dont know how to use Excel to solve these problems.

Think of it not as using Excel Solver, but as using the Marquart-Levenberg
algorithm.

choose three cells to contain the coefficients, and set the initial values to
something like 2300, 0, 0 (constant term, coeff of price, coeff of price squared).

Set up a column with the prices, next to it a column with the corresponding
sales, next to that a column with the predicted sales at the price in the first
column derived from the quadratic using the coefficients in the cells set up
to hold them earlier.

Now you need a new column with the square of the difference between the
sales and predicted sales for each of the prices in the left most column.

Now you need a cell containing the sum of the last column defined above.

Open solver, set to minimise, set cells to change to be those containing the
coefficients, then hit the solve button.

What you are asking Solver to minimise is the sum of the square differences
between sales and the models predicted sales at the same price.

It is easier to do than describe :mad:

RonL