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Jim\'s Camera shop sells two high-end cameras, the Sky Eagle and Horizon. The de

ID: 2949406 • Letter: J

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 = 230 - 0.5PS + 0.38PH

DH = 260 + 0.1Ps - 0.62PH

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:_______

Explanation / Answer

Answers:

Selling price of the Sky Eagle (Ps): 396.26

Selling price of the Horizon (PH): 363.02

Revenue: 176062.5

Solution:

given,

demand for sky eagle camera Ds = 230 - 0.5PS + 0.38PH

demand for horizon camera DH = 260 + 0.1Ps - 0.62PH

// revenue = demand times price

since we have 2 products revenue = Ds * Ps + DH * PH

revenue = Ps(230 - 0.5PS + 0.38PH) + PH(260 + 0.1Ps - 0.62PH)

= 230Ps - 0.5(Ps)^2 + 0.38PH.Ps + 260PH + 0.1Ps.PH - 0.62(PH)^2

simplifing the equation by derivating twice (once with respective to Ps and again with respective to PH)

dR / dPs = 230 - 1(Ps) + 0.358PH + 0 + 0.1PH - 0

= 230 - 1(Ps) + 0.458PH ....................let this be equation 1

dR / dPH = 0 - 0 + 0.38 Ps + 260 + 0.1Ps - 1.24 PH

= 260 + 0.48 Ps - 1.24 PH .........................let this be equation 2

solve equation 1 and 2 to get the unknown variables

1 x equation (1) = 230 - 1(Ps) + 0.458PH

2.084 x equation (2) = 541.84 +1(Ps) - 2.58416PH

by adding we get,

= 771.84 - 2.12616 PH = 0 // dR/ dPs = dR/ dPH = 0

= PH = 771.84/2.12616 = 363.0207

substituting PH in one of the equations to get Ps

= 230 - 1(Ps) + 0.458(363.0207) = 0

= 230 - Ps = - 166.2634806

= - Ps = - 166.2634806 - 230

= Ps = 396.2634

Revenue = 230Ps - 0.5(Ps)^2 + 0.38PH.Ps + 260PH + 0.1Ps.PH - 0.62(PH)^2

= 176062.5

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