Help! "An artist can sell 20 copies of a painting at $100 each..."

[lauranne]

here is the problem I need help with.

An artist can sell 20 copies of a painting at $100 each, but for each additional copy she makes, the value of each painting will go down by a dollar. Thus, if 22 copies are made, each will sell for $98. How many copies should she make to maximize her revenue?

 

[HallsofIvy]

here is the problem I need help with.

An artist can sell 20 copies of a painting at $100 each, but for each additional copy she makes, the value of each painting will go down by a dollar. Thus, if 22 copies are made, each will sell for $98. How many copies should she make to maximize her revenue?

For the first 20 copies she will earn 20(100)= $2000. Let x be the number of copies she sells beyond 20. For the first she will get $100- 1, for the second, 100- 2, etc. until for the "x"th copy she gets 100- x painting: she makes a total of $2000+ (100- 1)+ (100- 2)+ ...+ (100- x) where the "100- i" has x terms. That is the same as 2000+ (100+ 100+ ...+ 100)- (1+ 2+ 3+ ...+ x). 100 added to itself x times is 100x, of course. And it is a standard formula that 1+ 2+ 3+ ...+ x= x(x+1)/2. So the total amount of money she makes for selling x paintings (with x> 20) is 2000+ 100x+ x(x+1)/2.

Multiply that out and you get a quadratic formula for her earnings. You can find the value that makes her earnings a maximum by completing the square.

 

[lookagain]

For the first 20 copies she will earn 20(100)= $2000. Let x be the number of copies she sells beyond 20. For the first she will get $100- 1, for the second, 100- 2, etc. until for the "x"th copy she gets 100- x painting: she makes a total of $2000+ (100- 1)+ (100- 2)+ ...+ (100- x) where the "100- i" has x terms. That is the same as 2000+ (100+ 100+ ...+ 100)- (1+ 2+ 3+ ...+ x). 100 added to itself x times is 100x, of course. And it is a standard formula that 1+ 2+ 3+ ...+ x= x(x+1)/2. So the total amount of money she makes for selling x paintings (with x> 20) is 2000+ 100x+ x(x+1)/2.

Multiply that out and you get a quadratic formula for her earnings. You can find the value that makes her earnings a maximum by completing the square.

. HallsofIvy, my concern is that you misunderstood lauranne's exercise. Let x = the number of copies beyond 20. For each increase of one copy, the expression for the total number of copies is \(\displaystyle \bigg(20 + x\bigg). \ \ \) And as the number of copies increases by 1, the number of dollars that each copy is worth drops by $1. The expression for the value of each copy is \(\displaystyle \ \bigg($100 - ($1)x\bigg). \ \ \) I will drop the dollar sign symbols in simplifying the revenue function expression: \(\displaystyle \bigg(20 + x \bigg)\bigg(100 - x\bigg). \ \ \) As this is a calculus optimization problem, then lauranne would be expected to either FOIL it and take the derivative, or use the product rule for derivatives on it. Set the derivative equal to 0, and solve for the number of copies that must be added to 20.

 

[JeffM]

. HallsofIvy, my concern is that you misunderstood lauranne's exercise. Let x = the number of copies beyond 20. For each increase of one copy, the expression for the total number of copies is \(\displaystyle \bigg(20 + x\bigg). \ \ \) And as the number of copies increases by 1, the number of dollars that each copy is worth drops by $1. The expression for the value of each copy is \(\displaystyle \ \bigg($100 - ($1)x\bigg). \ \ \) I will drop the dollar sign symbols in simplifying the revenue function expression: \(\displaystyle \bigg(20 + x \bigg)\bigg(100 - x\bigg). \ \ \) As this is a calculus optimization problem, then lauranne would be expected to either FOIL it and take the derivative, or use the product rule for derivatives on it. Set the derivative equal to 0, and solve for the number of copies that must be added to 20.

You have ignored the boundary condition implicit in the problem. Students need to be aware that calculating a derivative and setting it to zero may not give the correct optimum when boundary conditions are involved (although in any particular case it may give the correct solution). It is easier to deal with the boundary condition and easier to see that it must be dealt with by setting x = the number of paintings sold.
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\(\displaystyle the\ value\ per\ copy = 100\ for\ 1 \le x \le 20;and\)
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\(\displaystyle the\ value\ per\ copy = 100 - (x - 20) = 120 - x\ for\ x > 20.\)
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Now \(\displaystyle revenue = 100x\ for\ 1 \le x \le 20\); and
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\(\displaystyle revenue = x(120 - x)\ for\ x > 20 \).
It is not sufficient to differentiate, etc. The boundary condition of x = 20 must be examined separately to be sure of finding the maximum.

 

[lookagain]

You have ignored the boundary condition implicit in the problem. Students need to be aware that calculating a derivative and setting it to zero may not give the correct optimum when boundary conditions are involved (although in any particular case it may give the correct solution). It is easier to deal with the boundary condition and easier to see that it must be dealt with by setting x = the number of paintings sold.
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>> > \(\displaystyle the\ value\ per\ copy = 100\ for\ 1 \le x \le 20;\) < < <

That doesn't follow. The problem makes no mention of being able to sell under 20 copies for any amount of money.

An artist can sell 20 copies of a painting at $100 each, but for each additional copy she makes, the value of each painting will go down by a dollar.

 

[JeffM]

That doesn't follow. The problem makes no mention of being able to sell under 20 copies for any amount of money.

Perhaps there is a discontinuity at x = 20, with revenue zero at all lower sales quantities: the problem is silent on that point. However, the problem certainly does leave open the possibility of selling 20 copies.

An artist can sell 20 copies of a painting at $100 each

If 20 copies is the sales quantity that maximizes revenue, it may not be a point where your function has a derivative of zero, particularly if it is discontinuous there. It is derelict in my opinion not to alert students to the importance of considering boundary conditions when searching for extrema.