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 = 220 - 0.5 P_s + 0.15 P_H

D_H = 275 + 0.2 P_s - 0.52 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(275 - 0.2 Ps - 0.52 PH) + Ps(220 - 0.5 Ps + 0.15 PH)
(ii)	PsDs - PHDH = Ps(220 - 0.5 Ps + 0.15 PH) - PH(275 - 0.2 Ps - 0.52 PH)
(iii)	PsDs + PHDH = Ps(220 - 0.5 Ps + 0.15 PH) + PH(275 + 0.2 Ps - 0.52 PH)
(iv)	PsDs - PHDH = Ps(220 + 0.5 Ps + 0.15 PH) - PH(275 - 0.2 Ps - 0.52 PH)

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

Find the prices that maximize revenue.

Do not round intermediate calculations. If required, round your answers to three decimal places.

Optimal Solution:

A) Selling price of the Sky Eagle (P_s): $________

B) Selling price of the Horizon (P_H): $________

C) Total Revenue: $___________

Given that Jim's camera. shop sells two high- End. DS = demand. the sky Eagle. PS as the selling price of the sky Eagle 014 is the demand for the Horizon and PH is de sellings price 8 to Horizon, DS = 2 20 - 0.5 PS 10. 15 PH DH = 275 + 0.2PS - 0.52 PH - ) we know that Revenue = DS. PS + DH-PH PS DS + PHD H = PS ( 2 2 0 - 0 . 5 P s + 0.15 PH ) + PH (275- +0.2Ps - 0.52PH) - so Equation with Respect 9 Ps and PH option ( ni ]\f297 -0 . 9795 PH = 0 PH = 297 0 . 7795 PH : $381 . 01 2 20 - 0.8Ps + 0 . 15 ( 3 2 1 . 01 ) 20 + 0.8 Ps = 277-15 PS = $ 346. 43 PH & PS value in Equation (2 ) R = 2 20 ( 346 - 43 ) - 0.5 ( 346 . 43 ) + 0.15 ( 381- 01) (346.43) + 275- ( 3 8 1 . 01 ) + 0 - 2 ( 3 4 6 . 4 3 ) ( 38 1 . 01 ) - 0:52 ( 38 1 . 01 ) 2. R 2 7 6 2 1 4 . 6 - 60, 0 06 . 87 + 19, 798 . 9 + 104, 727.7 + 261 398 -6 - 75; 487- 6. R = $ 91,695. 33