Click below to see ALL problems on Quadratic EquationsInquiry 1110766: The price p and also the quantity x marketed of a details product obey the demand equation listed below.x equates to negative 2 p plus 40 comma 0 less than or equates to p less than or amounts to 20x=−2p+40, 0≤p≤20 (a)Expush the revenue R as a duty of x.(b)What is the revenue if 44 devices are sold?(c)What amount x maximizes revenue? What is the maximum revenue?(d)What price must the firm charge to maximize revenue?(e)What price must the firm charge to earn at least $192 in revenue? Answer by Theo(11534)
(Sjust how Source): You deserve to put this solution on YOUR website! i think that what you wanted to say is:price p and also quantity x obey the demand also equation below:x = -2p + 40, wbelow 0 0 your revenue equation would be r = price * quantity = p * x.r is the revenuex is the number of units offered.p is the price per unit.the equation is r = p * xyou are given that x = -2p + 40solve for p to acquire p = 20 - .5xin the revenue equation of r = p * x, rearea p through 20 - .5x to get:r = (20 - .5x) * xsimplify to get r = 20x - .5x^2.set the revenue equal to 0 and also you obtain 0 = 20x - .5x^2rearrange the terms to obtain this equation in standard quadratic create to get: -.5x^2 + 20x = 0in this form:a = coeffective of x^2 term = -.5b = coreliable of x term = 20c = continuous term = 0.the maximum worth of this quadratic equation is once x = -b/2a.deal with for x to obtain x = -20 / -1 = 20.as soon as x = 20, the original equation of r = 20x - .5x^2 becomes r = 20*20 - .5*20^2 which becomes r = 400 - 200 which outcomes in r = 200.your maximum revenue is 200 when x = 20since r = x * p, then you have the right to settle for p to acquire p - r/x, which becomes:200/20 = 10the price per shirt would certainly need to be 10 in order to obtain a revenue of 200 as soon as the amount sold is 20.20 * 10 = 200you were previously given that x = -2p + 40when x = 20, the equation becomes 20 = -2p + 40subtract 40 from both sides to get -20 = -2pdivide both sides by -2 to gain p = 10.the price of 10 as soon as the amount marketed is 20 confirms the solution that tells you that the maximum revenue is 200 when the quantity offered is 20.answers to your questions are shown below:(a)Express the revenue R as a role of x.r = 20x - .5x^2(b)What is the revenue if 44 systems are sold?when x = 44, r = 20x - .5x^2 becomes:r = 20*44 - .5 * (44)^2 which becomes:r = -88because r = p * x, p would have had to be -2.this solution is invalid, because your trouble proclaimed that 0 therefore 44 units could not be offered, given that, by the equation, x = 44 outcomes in p = -2 which is out of the variety of the permissible worths of p.if you go ago to the original equation that says x = -2p + 40, then:as soon as p = 0, x = 40 and once p = 20, x = 0therefore 0 x = 44 is out of the permissible array of of both p and x.(c)What amount x maximizes revenue? What is the maximum revenue?as soon as x = 20, r = 200.this is as soon as the maximum revenue occurs.(d)What price should the company charge to maximize revenue?they must charge 10 dollars per unit.(e)What price should the firm charge to earn at least $192 in revenue?your equation is r = 20x - .5x^2if you desire the revenue to be higher than 192, then r should be > 192.this means that 20x - .5x^2 have to be better than 192.your equation becomes an inehigh quality that states 20x - .5x^2 > 192if you subtract 192 from both side of this inehigh quality, you get:20x - .5x^2 - 192 > 0to uncover the 0 points, collection 20x - .5x^2 - 192 = 0 and also solve for x.you will gain x = 16 or x = 24.that"s as soon as 20x - .5x^2 - 192 = 0given that 20x - .5x^2 represents the revenue, then that"s as soon as the revenue - 192 = 0 which indicates that"s as soon as the revenue = 192, bereason, when you include 192 to both sides of 20x - .5x^2 - 192 = 0, you gain 20x - .5x^2 = 192, which is the exact same as saying revenue = 192.so when x = 16 or once x = 24, the revenue is equal to 192.to discover out what interval the revenue is greater than 192, then pick worths much less than 16, higher than 24, between 16 and 24.you will find that the revenue is better than 192 once the quantity of x is > 16 and also the graph of the original revenue equation reflects that to be true.here"s the graph.