Question

(1 point) A manufacture has been selling 1100 television sets a week at $450 each. A market survey indicates that for each $10 rebate offered to a buyer, the number of sets sold will increase by 100 per week. a) Find the function representing the demand p(x), where x is the number of the television sets sold per week and p(x) is the corresponding price. P(x) = b) How large rebate should the company offer to a buyer, order to maximize its revenue? dollars c) If the weekly cost function is 82500+ 150x, how should it set the size of the rebate to maximize its profit? dollars

Answer 1

1 Given demand function when for 1100 TV's te pel100) 450 If debate $ 10 is given then sale in coeuses by 100 per week A fos peice - deceases of the locato per unit the sele will increase by 100 unit per week ? Let x be the number of unit sold per week then is the increase in the number of unit Sold P(0 = 450 -1 (x-1100) 450 - 2 +11o X-lloc jo 10 560 2 Revenue Ro) = x-P(X) (560-ge) = 5.60%-x 10 - Ro) -560 5 when Revenue is maximum, then Row=0 560 x=560xS 2800 Here RW) 70 for o <x< 2800 & Row co for a7 2800 - Imaximum occuses at 2= 2800. P (2800) = 560 2800 = 560-280 280 Hence, the zebate to maximize the be 450-280 $170 revenue fit should

Cost function is Cox)= Cx) = 82500 + 150x Hence, profit function Por= Row - ((x) Daces = (5602 - x? ) - (82500 +150x) P(1) - -0² 10 + 4100 -82500 marginal profit P'low a -x + 410 5 Put P'(x) 20 = -x +410 =0 2050 = x = 41045 = x - 2050) P620.50) = Sъо تیانى 2050 10 560-205 355 Thus, the rebate to maximize the revenue should be 450-355 = $95 1$95