Jim\'s Camera shop sells two high-end cameras, the Sky Eagle and Horizon. The de

Question

Jim's Camera shop sells two high-end cameras, the Sky Eagle and Horizon. The demand for these two cameras are as follows (D_S = demand for the Sky Eagle, P_s is the selling price of the Sky Eagle, D_H is the demand for the Horizon and P_H is the selling price of the Horizon): D_s = 222 - O.6P_s + 0.35PH D_H = 270 + 0.1P_s - 0.64P_H The store wishes to determine the selling price that maximizes revenue for these two products. Develop the revenue function for these two models. Choose the correct answer below. (i) P_5 D_5 + P_HD_H = P_H(270 - 0.1P_5 - 0.64P_H) + P_5(222 - 0.6P_5 + 0.35PH) (ii) P_5D_5 - P_HD_H = P_5(222 - 0.6P_5 + 0.35P_H) - P_H(270 - 0.1P_5 - 0.64P_H) (III) P_s D_s + P_HD_H = P_5(222 = 0.6P_5 + 0.35P_H) + P_H(270 + 0.1P_S5 = 0.64P_H) (iv) P_5D_5 - P_HD_H = P_5(222 + 0.6P_5 + 0.35P_H) - P_H(270 - 0.1P_5 - 0.64P_H) Find the prices that maximize revenue. If required, round your answers to two decimal places. Optimal Solution: Selling price of the Sky Eagle (P_s): $ Selling price of the Horizon (P_H): $ Revenue: $

Explanation / Answer

Demand Function for both type of cameras

Ds = 222 - 0.6 ps + 0.35 ph

DH = 270 + 0.1Ps - 0.64 Ph

Revenue = Ds * ps + Dh * ph

= (222 - 0.6 ps + 0.35 ph )ps + ( (270 + 0.1Ps - 0.64 Ph  ) ph

Option C is correct.

let say ps= x and py = y

so Total reveunue = ( 222- 0.6 x + 0.35 y) x + (270 + 0.1 x - 0.64y) y

TR = 222 x - 0.6 x2 + 0.35 xy + 270 y + 0.1 xy - 0.64 y2

TR = 222x + 270 y  - 0.6 x2 - 0.64 y2 + 0.45 xy

so difeerentiatling TR with respect to x and y will give maximum revnue results

d(TR)/dx = 222 - 1.2 x + 0.45 y

d(TR)/ dy = 270 - 1.28 y + 0.45 x

for optimum prices d(TR)/dx = 0 = d(TR)/ dy

so 222 - 1.2 x + 0.45 y = 0 = 270 - 1.28 y + 0.45 x = 0

by solving these 2 linear equations

x = 304.21 and y = 317.88

by putting value of x and y in TR equation we will get total reveenue

And Total revenue = $76681.48

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