the owner of an aquatic store used 17 1/2 gallons of water to fill aquariums. he puts 5 5/6 gallons of water in each aquarium. h
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Answer:
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.
<u>Mathematically</u>
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
<u>Graphically</u>
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.
<u>Maximum Profit</u>
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
<u>Profit = $1702</u>
ur answer is-
<h2>$12.500</h2>- Thanks
- Hope it helps :).