Pumpkin PriceDate: 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 http://mathforum.org/dr.math/ 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 http://mathforum.org/dr.math/ |
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