Using my software, I found that
is and excellent match.
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?
Of course there is, it is just extremely messy which is why I used software.Originally Posted by askmemath
Click here to infect your computer
This is not a good predictor of sales from price.Originally Posted by ThePerfectHacker
A better model can be obtained using the Excel Solver, this givesCode:>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 solution:
This gives:
A plot comparing this model with the estimated sales is shown in the attachmentCode:> >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 >
RonL
With the Excel solver we are solving a non-linear regression problem, whichOriginally Posted by askmemath
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
Think of it not as using Excel Solver, but as using the Marquart-LevenbergOriginally Posted by askmemath
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
RonL