Maximization problem

Question

An evening mathematics course at BHS has a maximum capacity of 100 students. When the tuition for enrolling in the course is $80 all of the 100 seats are filled. Experience shows that for each $5 per month increase in tuition, five fewer students enroll. BHS absorbs the cost of supplies and instruction, which comes to $20 per student. What tuition should be charged to maximize the profit for offering this course?

Remember that profit equals revenue minus cost.

Answer 1

I believe Mark M. is correct...R(x)=(80+5x)*(100-5x) where 80+5x = tuition cost and 100-5x=# of students enrolled. If your trying to find the max. profit, you'll need to factor in the cost (i.e. P(x)=R(x)-C(x)). The cost equation in this case is $20 * (# of Students Enrolled). As stated before, the # of Students Enrolled = 100-5x.

Now, you can simplify P(x).

P(x)=R(x)-C(x)=(80+5x)*(100-5x)-20*(100-5x)

P(x)=8000-400x+500x-25x²-2000+100x

P(x)=-25x²+200x+6000

Because the x² coefficient is negative, P(x) is going to be concave down. So, if you derive P(x) and set that derivative equal to 0, you'll find the x value that corresponds to the maximum profit.

dP/dx = -50x+200=0 --> x=4

When x=4, the profit is maximized. In order to figure out how much tuition should be charged, just plug x=4 into the equation that corresponds with the tuition price (80+5x). So, maximum tuition price = 80+5*4=80+20=$100. $100 is the price that corresponds with the maximum profit.

Answer 2

R(x) = (80 + 5x)(100 - 5x)

Can you convert to vertex form and answer?