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 (DS = demand for the Sky Eagle, Ps is the selling price of the Sky Eagle, DH is the demand for the Horizon and PH is the selling price of the Horizon):

Ds = 215 - 0.5PS + 0.15PH

DH = 260 + 0.2Ps - 0.45PH

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.

- 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 (Ps): $ ____________________

Selling price of the Horizon (PH): $ __________________-

Revenue: $______________

(i) PsDs + PHDH = PH(260 - 0.2Ps - 0.45PH) + Ps(215 - 0.5Ps + 0.15PH) (ii) PsDs - PHDH = Ps(215 - 0.5Ps + 0.15PH) - PH(260 - 0.2Ps - 0.45PH) (iii) PsDs + PHDH = Ps(215 - 0.5Ps + 0.15PH) + PH(260 + 0.2Ps - 0.45PH) (iv) PsDs - PHDH = Ps(215 + 0.5Ps + 0.15PH) - PH(260 - 0.2Ps - 0.45PH)

Explanation / Answer

Revenue is maximized when we find total revenue and finds it derivative

Revenue = PsDs + PHDH

= Ps(215 - 0.5PS + 0.15PH) + PH(260 + 0.2Ps - 0.45PH)

Correct match is found in (iii)

This makes

TR = 215PS - 0.5PS^2 + 0.15PHPS + 260PH + 0.2PSPH - 0.45PH^2

= 215PS - 0.5PS^2 + 260PH - 0.45PH^2 + 0.35PSPH

Maximize TR wrt PS and PH partially

215 - PS + 0.35PH = 0

260 - 0.9PH + 0.35PS = 0

Solve them to get

Selling price of the Sky Eagle (Ps): $365.92

Selling price of the Horizon (PH): $431.19

Revenue: $95390.68

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