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Pumpkin Price

Date: 06/28/2001 at 00:34:46
From: Sharmila
Subject: Quadratic Equations

Peter Peter has 40 pumpkins to sell. He will sell them all if he sets 
his price at $1.00 each, but will sell one fewer pumpkin for each 
5-cent increase in price. What price should he charge in order to 
maximize his earnings? [Show work algebraically.]     

I have tried everything on this question. I even tried asking other 
people, but nobody can figure it out. I've let x equal the number of 
cents, etc. I used the 'completing the square' formula, which you 
would usually use on a question like this, but I still have had no 
luck. Help!

Date: 06/28/2001 at 16:04:12
From: Doctor Rob
Subject: Re: Quadratic Equations

Thanks for writing to Ask Dr. Math, Sharmila.

Let x be the price for which he sells his pumpkins in cents. The
number he sells is then 40 - (x-100)/5 = 60 - x/5. Then y, the amount
of money he earns, is the number of pumpkins time the price, which
comes out to be

   y = x*(60-x/5),
     = 60*x - x^2/5,
     = (300*x - x^2)/5.

Now complete the square on x in the quantity in the numerator, and
find the value of x that makes the numerator the largest.

- Doctor Rob, The Math Forum   

Date: 06/28/2001 at 16:08:34
From: Doctor Jaffee
Subject: Re: Quadratic Equations

Hi Sharmila,

If x = the number of times he increases the price $.05, then the cost 
of one pumpkin will be 1 + .05x, but the number of pumpkins he will 
sell at that price will 40 - x.

The amount of money he will earn is the product of these two.

So, multiply them together and put your answer in the form

   y = ax^2 + bx + c, 

where y is the amount of earnings.

Recall that the graph of a quadratic is a parabola, and that the 
maximum point when a is negative is at the vertex or one of the end 
points (x can't be smaller than 0, nor larger than 40).

Also, recall that the x value at the vertex of any parabola is -b/2a.

Give it a try and if you want to check your solution with me, write 
back. If you are having difficulties, let me know and show me what you 
have done so far, and I'll try to help you some more.

Good luck.

- Doctor Jaffee, The Math Forum   
Associated Topics:
High School Basic Algebra

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