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 = 215 - 0.6 P_s + 0.25 P_H

D_H = 270 + 0.1 P_s - 0.55 P_H

The store wishes to determine the selling price that maximizes revenue for these two products. Select the revenue function for these two models. Choose the correct answer below.

(i)	PsDs + PHDH = PH(270 - 0.1 Ps - 0.55 PH) + Ps(215 - 0.6 Ps + 0.25 PH)
(ii)	PsDs - PHDH = Ps(215 - 0.6 Ps + 0.25 PH) - PH(270 - 0.1 Ps - 0.55 PH)
(iii)	PsDs + PHDH = Ps(215 - 0.6 Ps + 0.25 PH) + PH(270 + 0.1 Ps - 0.55 PH)
(iv)	PsDs - PHDH = Ps(215 + 0.6 Ps + 0.25 PH) - PH(270 - 0.1 Ps - 0.55 PH)

- Select your answer -Option (i)Option (ii)Option (iii)Option (iv)Item 1

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: $

Answer 1

Solution:-

Given

DS = 215 - 0.6 (PS) + 0.25 (PH)

DH = 270 + 0.1 (PS) - 0.55 (PH)

Revenue = DS . PS + DH . PH

= (215 - 0.6 (PS) + 0.25 (PH)) (PS) + (270 + 0.1 (PS) - 0.55 (PH)) (PH)

R = Revenue =

To find maximising Revenue, we have to differentiate above equation w r t to PS & PH

= 215 - 1. 2(PS) + 0.35 (PH)

= 0.35 (PS) + 270 - 1.1 (PH)

Now we have to equate both

solving

75.25 - 0.42 (PS) + 0.1225 (PH) = 0

324 + 0.42 (PS) - 1.32 (PH) = 0

By solving 2 equations

399.25 - 1.1975 (PH) = 0

PH = 333.40

PS 215 - 1.2 (PS) + 0.35 (PH) = 0

215 - 1.2 (PS) + 0.35 (333.40) = 0

1.2 PS = 215 + 116.69

PS = 276.41

Revenue = 215 (PS) - 0.6 (PS)2 + 0.25 (PH) (PS) + 270 (PH) + 0.1 (PH)(PS)- 0.55 (PH)2

= 215 (276.41) - 0.6 (276.41)2 + 0.25 (333.40)(276.41) + 270 (333.40) + 0.1 (333.40)(276.41) - 0.55 (333.40)2

= 74723.38