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Let, x = number of $10 rebates given. Now, at x rebated the price is, Price = $450-$10x And the quantity sold is (1000+100x). Now,
Let, x = number of $10 rebates given. Now, at x rebated the price is, Price = $450-$10x And the quantity sold is (1000+100x). Now, Quantity = 1000+100x X = (Quantity-1000)/100 Putting it in previous equation gives, Price = $450-$10*(Quantity-1000)/100 = $450-0.1*Quantity+100 = $550- 0.1*Quantity This is the price determined from the demand function. Now, the revenue function is, Revenue = Price*Quantity = ($450-$10x)*(1000+100x). = -1000x2+35000x+450000 Now at the maximum point, (d/dx) Revenue = 0 -2*1000x +35000 = 0 X = 35000/2000 = 17.5 So the total rebate = $10x = $10*17.5 = $175. Let, x = number of $10 rebates given. Now, at x rebated the price is, Price = $450-$10x And the quantity sold is (1000+100x). Now, Quantity = 1000+100x X = (Quantity-1000)/100 Putting it in previous equation gives, Price = $450-$10*(Quantity-1000)/100 = $450-0.1*Quantity+100 = $550- 0.1*Quantity This is the price determined from the demand function. Now, the revenue function is, Revenue = Price*Quantity = ($450-$10x)*(1000+100x). = -1000x2+35000x+450000 Now at the maximum point, (d/dx) Revenue = 0 -2*1000x +35000 = 0 X = 35000/2000 = 17.5 So the total rebate = $10x = $10*17.5 = $175. Here x = 17.5 at the maximum revenue point so putting them in proper equations give, Quantity to maximize revenue = 1000+100x = 1000+100*17.5 = 2750 Price to maximize revenue = $550- 0.1*Quantity = $550- 0.1*2750 = $275 Maximum revenue = -1000x2+35000x+450000 = -1000*17.52+35000*17.5+450000 = $756,250.00 Let, x = number of $10 rebates given. Now, at x rebated the price is, Price = $450-$10x And the quantity sold is (1000+100x). Now, Quantity = 1000+100x X = (Quantity-1000)/100 Putting it in previous equation gives, Price = $450-$10*(Quantity-1000)/100 = $450-0.1*Quantity+100 = $550- 0.1*Quantity This is the price determined from the demand function. Now, the revenue function is, Revenue = Price*Quantity = ($450-$10x)*(1000+100x). = -1000x2+35000x+450000 Now at the maximum point, (d/dx) Revenue = 0 -2*1000x +35000 = 0 X = 35000/2000 = 17.5 So the total rebate = $10x = $10*17.5 = $175. Here x = 17.5 at the maximum revenue point so putting them in proper equations give, Quantity to maximize revenue = 1000+100x = 1000+100*17.5 = 2750 Price to maximize revenue = $550- 0.1*Quantity = $550- 0.1*2750 = $275 Maximum revenue = -1000x2+35000x+450000 = -1000*17.52+35000*17.5+450000 = $756,250.00 Let, x = number of $10 rebates given. Now, at x rebated the price is, Price = $450-$10x And the quantity sold is (1000+100x). Now, Quantity = 1000+100x X = (Quantity-1000)/100 Putting it in previous equation gives, Price = $450-$10*(Quantity-1000)/100 = $450-0.1*Quantity+100 = $550- 0.1*Quantity This is the price determined from the demand function. Now, the revenue function is, Revenue = Price*Quantity = ($450-$10x)*(1000+100x). = -1000x2+35000x+450000 Now at the maximum point, (d/dx) Revenue = 0 -2*1000x +35000 = 0 X = 35000/2000 = 17.5 So the total rebate = $10x = $10*17.5 = $175. Here x = 17.5 at the maximum revenue point so putting them in proper equations give, Quantity to maximize revenue = 1000+100x = 1000+100*17.5 = 2750 Price to maximize revenue = $550- 0.1*Quantity = $550- 0.1*2750 = $275 Maximum revenue = -1000x2+35000x+450000 = -1000*17.52+35000*17.5+450000 = $756,250.00 Let, x = number of $10 rebates given. Now, at x rebated the price is, Price = $450-$10x And the quantity sold is (1000+100x). Now, Quantity = 1000+100x X = (Quantity-1000)/100 Putting it in previous equation gives, Price = $450-$10*(Quantity-1000)/100 = $450-0.1*Quantity+100 = $550- 0.1*Quantity This is the price determined from the demand function. Now, the revenue function is, Revenue = Price*Quantity = ($450-$10x)*(1000+100x). = -1000x2+35000x+450000 Now at the maximum point, (d/dx) Revenue = 0 -2*1000x +35000 = 0 X = 35000/2000 = 17.5 So the total rebate = $10x = $10*17.5 = $175. Here x = 17.5 at the maximum revenue point so putting them in proper equations give, Quantity to maximize revenue = 1000+100x = 1000+100*17.5 = 2750 Price to maximize revenue = $550- 0.1*Quantity = $550- 0.1*2750 = $275 Maximum revenue = -1000x2+35000x+450000 = -1000*17.52+35000*17.5+450000 = $756,250.00
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