How Do You Solve a Max Min Quadratic Problem in Economics?

[aisha]

I used to know how to do this but I can't remember can someone please help me out?

An amusement park charges $8 admission and average of 2000 visitors per day. A survey shows that, for each 1$ increases in the admission cost, 100 fewer ppl would visit the park.

Write an equation to express the revenue R(X) dollars, in terms of a price increase of x dollars

find the coordinates of the maximum point of this function--> I think for this you complete the square and the coordinates is the vertex.

what admission cost gives you the maximum revenue--> is this the x value?

how many visitors give the maximum revenue? is this the y value?

please help me I am a little lost thanks

 

[BobG]

I used to know how to do this but I can't remember can someone please help me out?

An amusement park charges $8 admission and average of 2000 visitors per day. A survey shows that, for each 1$ increases in the admission cost, 100 fewer ppl would visit the park.

Write an equation to express the revenue R(X) dollars, in terms of a price increase of x dollars

find the coordinates of the maximum point of this function--> I think for this you complete the square and the coordinates is the vertex.

what admission cost gives you the maximum revenue--> is this the x value?

how many visitors give the maximum revenue? is this the y value?

please help me I am a little lost thanks

Set up an equation that relates price and visitors to revenue (8 is your base price and 2000 is your base visitors, so as price increases by a certain amount, how much does your number of visitors decrease)

Take the derivative of the equation.

Set the derivative to zero and solve for your variable. (First derivative test). The solution(s) will usually be either a local maximum or local minimum. Do the second derivative test by plugging the solution into the second derivative (>0 means min.; <0 means max; 0 or does not exist means test fails and you have to do a couple trial and error test cases).

In this case, you only have one solution that makes the derivative equal to zero. If you reduce the price to zero, you'll have no revenue (derivative is not zero, since if you start paying to come, you'll have plenty of visitors). If you increase the price too far, you'll have no visitors, hence no revenue (derivative is not zero on this end either - in fact, you can have negative visitors - the price is so high that even the people who illegally jumped the fence to get in want a refund). So, it should be obvious that any critical point (such as a derivative equal to zero) in between will be your max.

 

[aisha]

Got it don't know if its right?

can someone check this well i got the eqn to be
R=(2000-100x)(8+x)

after completing the square i got vertex of (6,19600) so this means that the admission should be increase by 6 dollars to get max revenue
the number of visitors that give the max revenue are 19600 is this right?
thanks for ur help every one

 

[BobG]

Yes, a $6 increase gets you revenue of $19,600 (1400 visitors)