Can anyone please help? I keep posting this same question and getting incorrect

Question

Can anyone please help? I keep posting this same question and getting incorrect answers.

A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x-74,000+ 80x and p (x)-250- , 0 5000. 20 (A) Find the maximum revenue. (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set (C) If the government decides to tax the company $4 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for eaclh set? A) The maximum revenue is S 312500.00 Type an integer or a decimal) (B) The maxim um profit is $1 Type integers or decimals.) when sets are manufactured and sold for $ each. Please don't forget to answer part c as well

Explanation / Answer

(A) The revenue equation is given by x.p(x) = x(250-x/20)

Now the maximum revenue is obtained by differentiating this:

R'(x) = 250-2x/20 = 0 => 250-x/10 = 0 => 2500 = x

R"(x) = -2/20 = -1/10<0 => R(x) is maximu at x=2500

So R(2500) = 2500(250-2500/20) = 2500(250-125) = 2500(125) = $312500

(B)The profit is given by P(x) = R(x) - C(x) = 250x-x2/20-74,000-80x = 170x-x2/20-74,000

Now P'(x) = 170-2x/20 = 0 => 170-x/10=0 => x=1700

P"(x) = -2/20=-1/10<0 => P(x) is maximum for x=1700

So P(1700) = 170(1700)-(1700)2/20 - 74,000 = 289000-144500-74000 = 70500

So the maximum profit is 7,0500 when 1700 televesion sets are sold

Cost for each is p(1700) = 250-1700/20 = 250-85 = $165 for each television set

(C) Let t(x) be the taxing function

Then t(x) = 4x

Then cost function is given by: c(x) = 4x+74000+8x = 74000+12x

So the profit function now becomes:

P(x) = R(x)-c(x) = 250x-x2/20-74000-12x = 238x-x2/20-74000

Now P'(x) = 238-2x/20 = 238-x/10 = 0 => x=2380

P"(x) = -2/20 = -1/10 <0 => profit is maximum for x=2380

Now P(2380) = 238(2380)-(2380)2/20-74000 =566440- 283220-74000=$209220

So the Maximum profit in this case is $209220 when 2380 television sets are sold

p(2380) = 250-2380/20 = 250-119= $131

The company should sell each televison set at the cost of $131 per televison set

A car rental agency.....

The income for one day at starting = 480(40) =$19200 per day

For each $1 increase we get fllowing equation:

y = (480-10x)(40+x)

This equation is quadratic and has a maximum.For this find zeros of y

y = 0 at x=48 and x=-40

Now mid point of these is :

x=48+(-40)/2 = 48-40/2 = 8/2 = 4

This means the rental rate should be 40+4 = $44 per day

Maximum income is (480-40)(44) = $19360

So the maximum income is $19360 when it charges $44 per day

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