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 = 222 - 0.6PS + 0.35PH

DH = 270 + 0.1Ps - 0.64PH

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(270 - 0.1Ps - 0.64PH) + Ps(222 - 0.6Ps + 0.35PH) (ii) PsDs - PHDH = Ps(222 - 0.6Ps + 0.35PH) - PH(270 - 0.1Ps - 0.64PH) (iii) PsDs + PHDH = Ps(222 - 0.6Ps + 0.35PH) + PH(270 + 0.1Ps - 0.64PH) (iv) PsDs - PHDH = Ps(222 + 0.6Ps + 0.35PH) - PH(270 - 0.1Ps - 0.64PH)

Explanation / Answer

Ds = 222 - 0.6PS + 0.35PH

DH = 270 + 0.1Ps - 0.64PH

Revenue = PsDs + PHDH

=Ps(222 - 0.6Ps + 0.35PH) + PH(270 + 0.1Ps - 0.64PH)

option (iii) is the correct answer

To maximise revenue,

derivative of revenue with respect to Ps = derivative of revenue with respect to PH = 0

Revenue, R = Ps(222 - 0.6Ps + 0.35PH) + PH(270 + 0.1Ps - 0.64PH)

derivative of Revenue with respect to Ps,

222 -1.2Ps + 0.45PH = 0 eq(1)

derivative of revenue with repsect to PH

270+ 0.45Ps - 1.28PH = 0 eq(2)

solving eq(1) and eq(2), we get

Ps = $ 304.21

PH = $ 317.88

substituting these values in revenue equation, we get

Revenue = $ 76681.48

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