These are the cost and revenue functions for a line of trumpets sold at a music store: R(x) = 76x – 0.25x2
C(x) = -7.75x + 5,312.5

Based on these functions, what is the maximum profit that can be made, to the nearest dollar?

A.
$1,469
B.
$1,702
C.
$3,375
D.
$5,776

168 trumpets for $1702

Step-by-step explanation:

Profit is the measure to be maximized.  We are given revenue and cost relationships as a function of units, x (trumpets).  Profit is the difference:

Profit = Revenue[R(x)] - Cost[C(x)]

Profit = (76x – 0.25x^2) - (-7.75x + 5,312.5)

Profit = 76x - 0.25x^2 + 7.75x - 5,312.5

Profit = 76x - 0.25x^2 + 7.75x - 5,312.5

Profit = - 0.25x^2 + 83.75x - 5312.5

At this point we can find the trumpets needed for maximum profit by either of two approaches:  algebraic and graphing.  I'll do both.

Mathematically

The first derivative will give us the slope of this function for any value of x.  The maximum will have a slope of zero (the curve changes direction at that point).  Take the first derivative and set that equal to 0 and solve for x.

First derivative:

d(Profit)/dx = - 2(0.25x) + 83.75

d(Profit)/dx = - 0.50x + 83.75

0 = - 0.50x + 83.75

0.50x = 83.75

x = 167.5 trumpets

Graphically

Plot the profit function and look for the maximum.  The graph is attached.  The maximum is 167.5 trumpets.

Round up or down to get a whole trumpet.  I'll go up:  168 trumpets.

Maximum Profit

Solve the profit equation for 168 trumpets:

Profit = - 0.25x^2 + 83.75x - 5312.5

Profit = - 0.25(168)^2 + 83.75(168) - 5312.5

Profit = $1702

5050

hope this !

do you have a picture of the table