Expert: Sherman D. - 1/18/2005

Question
Problem:   Please help, if you can

  p = -1/40x + 80


Express the revenue (R) as a function of x.

What is the revenue if 500 items are sold?

What quantity (x) maximizes the revenue?

What price should the company charge to maximize the revenue?

Tell what kind of symmetry the graph of the following relation would have:

 x = y^2 - 2

Answer
if by this you mean

P = (-1/40)x + 80
P = -((1/40)x - 80)

Assuming by this you mean "p" = profit, (1/40)x = Revenue, and 80 = cost.

and if so, then

P = -((1/40)x - 80)

x = -((1/40)P - 80)

-x = (1/40)P - 80

-x + 80 = (1/40)P

-40x + 3200 = P

R(x) = -40x + 3200

R(500) = -40(500) + 3200
R(500) = -2000 + 3200
R(500) = 1200

since this problem doesn't have a x^2 in the problem, the problem would be a linear function so therefore the quantity that maximizes the revenue would constintently continue to increase, so here is what i can do

R(x) = -40x + 3200
0 = -40x + 3200
-3200 = -40x
x = 80

so the answer would be x < 80, so since you can't have a negative quantity, the answer would be 0 < x < 80

If i have misunderstood you in anyway, just let me know.

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x = y^2 - 2

the graph has a horizontal even symmetry at y = 0.

Most of the time, when the highet power is an even number, the graph is even, and when the highest power is an odd number, the graph is odd.

Most even problems are in the form of y = ax^2 + bx + c and most odd problems are in the form of y = ax^3 + bx^2 + cx + d.