Question

Problem 13-5

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 = 230 - 0.5PS + 0.38PH

DH = 260 + 0.1Ps - 0.62PH

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)	PsDs + PHDH = PH(260 - 0.1Ps - 0.62PH) + Ps(230 - 0.5Ps + 0.38PH)
(ii)	PsDs - PHDH = Ps(230 - 0.5Ps + 0.38PH) - PH(260 - 0.1Ps - 0.62PH)
(iii)	PsDs + PHDH = Ps(230 - 0.5Ps + 0.38PH) + PH(260 + 0.1Ps - 0.62PH)
(iv)	PsDs - PHDH = Ps(230 + 0.5Ps + 0.38PH) - PH(260 - 0.1Ps - 0.62PH)

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

Find the prices that maximize revenue.

If required, round your answers to two decimal places.

Optimal Solution:

1. Selling price of the Sky Eagle (Ps): $

2. Selling price of the Horizon (PH): $

3. Revenue: $

Answer 1

Revenue = Price x Demand

So, Option III is correct : PsDs + PHDH = Ps(230 - 0.5Ps + 0.38PH) + PH(260 + 0.1Ps - 0.62PH)

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Revenue function (R) = Ps(230 - 0.5Ps + 0.38PH) + PH(260 + 0.1Ps - 0.62PH)

Take partial derivative w.r.t. Ps and equate it with zero

i.e. 230 - Ps + 0.38PH + 0 + 0.1PH - 0 = 0
or, Ps - 0.48PH = 230 -----(1)

Take partial derivative w.r.t. PH and equate it with zero

i.e. 0 - 0 + 0.38Ps + 260 + 0.1Ps - 1.24PH = 0
or, 1.24PH - 0.48Ps = 260 ---(2)

Solving (1) and (2) as simultaneous linear equations, we get Ps = $406.10 and PH = $366.88 and the Revenue becomes = Ps(230 - 0.5Ps + 0.38PH) + PH(260 + 0.1Ps - 0.62PH) = $94395.80