Calculus practice review problem: At $4 price per frozen yogurt, demand is 900 units.

[stevesmith99]

At a price of $4 per frozen yogurt, demand is 900 units. Every decrease of one dollar, the demand increasesby 200 units. What price maximizes revenue?

(a) 640(b) 850(c) 4.25(d) 8.50

The question is from a review packet I am working on. If anyone can help me figure out how to answer it I would really appreciate it, thanks.

 

[stapel]

At a price of $4 per frozen yogurt, demand is 900 units. Every decrease of one dollar, the demand increasesby 200 units. What price maximizes revenue?

(a) 640(b) 850(c) 4.25(d) 8.50

The question is from a review packet I am working on. If anyone can help me figure out how to answer it I would really appreciate it, thanks.

Use what you learned back in algebra. You are given that things change in terms of decreases of a dollar. You're needing to figure out how many decreases would be optimal. So pick a variable for this unknown value; say, "d" for "decreases".

Then, to figure out the pattern, try using actual numbers.

. . . . .zero decreases: price is $4 - 0*1, demand is 900 + 0*200, revenue is 900*$4

. . . . .one decrease: price is $4-1*1 = $3, demand is 900 + 1*200 = 1,100, revenue is 1,100*$3

. . . . .two decreases: price is $4-2*1 = $2, demand is 900 + 2*200 = 1,300, revenue is 1,300*$2

And so forth. What would be the pattern for "d" decreases? What "revenue" equation can you then create? What would be the maximizing value of "d"?

If you get stuck, please reply showing your work and reasoning so far. Thank you!

 

[HallsofIvy]

At a price of $4 per frozen yogurt, demand is 900 units. Every decrease of one dollar, the demand increasesby 200 units. What price maximizes revenue?

(a) 640(b) 850(c) 4.25(d) 8.50

The question is from a review packet I am working on. If anyone can help me figure out how to answer it I would really appreciate it, thanks.

As stapel suggested, this is more a review of algebra, in preparation for Calculus, than a review of Calculus.

Since, for a fixed decrease in the price, the demand increases by a fixed amount, the demand is a linear function of price with slope -200 units per dollar:
d= -200P+ C where C is a constant. When the price is P= 4, d= 900 so 900= -200(4)+ C. C= 900+ 800= 1700. d= 1700- 200P.

"Revenue" is the number sold times the price of each: R= (1700- 200P)(P)= 1700P- 200P^2. That is a parabola opening downward. To find the price that "maximizes revenue" we need to find P at the vertex of the parabola. We can do that by "completing the square" in R= -200(P^2- 8.5P).

 

[lookagain]

As stapel suggested, this is more a review of algebra, in preparation for Calculus, than a review of Calculus.

This is a typical type of an applied calculus course maximization problem for economics and social
sciences, for instance. The student is expected to set up a function, take the derivative, set that equal
to zero, solve that for any appropriate critical numbers, find out which critical number(s) maximize,
and then commonly substitute that number into the price expression if required. **

And, by algebra, when the revenue is written out as R(x) = ax^2 + bx + c, the x-value at the max is
given by -b/(2a), if someone wanted to check it another way.

________________________________________________________


Let x = the value that will maximize the revenue


(demand in units)*(price in dollars)

(900 . .... . . )*(4 . . . . . )

The above is what it starts out as.


Revenue = (900 + 200x)*(4 - 1x)

or

R(x) = 100(9 + 2x)(4 - x)


** - - See these steps as a general guide to continue.

 

[lookagain]

It's been about 4 days since the posting of the problem, so I wanted to show it carried forward:


Continuing from post #4:


R(x) = 100(9 + 2x)(4 - x)

R(x) = 100(36 - 9x + 8x - 2x^2)

R(x) = 100(36 - x - 2x^2)

R'(x) = 100(-1 - 4x)

100(-1 - 4x) = 0

-1 - 4x = 0

-4x = 1

x = -1/4 = -0.25

price = 4 - x

price = 4 - (-0.25)

= 4.25


$4.25 is the price that maximizes revenue.