Problem 13-5 Jim\'s Camera shop sells two high-end cameras, the Sky Eagle and Ho
ID: 334926 • Letter: P
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.
- 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: $
(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)Explanation / Answer
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