A company is willing to promote and sell two types of smart watches, X and Y. Th

Question

A company is willing to promote and sell two types of smart watches, X and Y. The demand for these two watches are as follows.

dx = -0.45px + 0.34py + 242

dy = 0.2 px -0.58py + 282

where, dx is the demand and px is the selling price of watch X respectively, dy is the demand and py is the selling price of watch Y respectively.

Company wishes to determine the selling price that maximizes revenue for these two products. Total production capacity for the two watches is limited to 5000 watches.

Formulate the problem to maximize the total revenue.

Explanation / Answer

Revenue = Price x quantity demanded

Revenue for smart watch X = (-0.45px + 0.34py + 242) * px

Revenue for smart watch X = -0.45px2 + 0.34pxpy + 242px

taking the first derivative of the revenue function

dpx/ dp= -0.90 px + 0.34py +242 - (1)

taking the second derivative

dpx2 /dp = -0.90

Thus the revenue function is maximized

Equating 1 equation to zero

-0.90 px + 0.34py +242 = 0

The demand for watch y is dy = 0.2 px -0.58py + 282

Revenue =  (0.2px -0.58py + 282)py

R = 0.2 pxpy - 0.58py2 + 282py

Taking the first derivative = 0.2px - 1.16py + 282

The second derivative gives = -1.16

Thus the revenue function is maximzied

Eqating the first derivative to zero

0.2px - 1.16py + 282 = 0 (1)

-0.90 px + 0.34py +242 = 0 (2)

Multiply equation 1 by 4.5

0.90px - 5.22py + 1269 = 0 ( 1)

-0.90x + 0.34py + 242 = 0 (2)

adding equation 1 and 2

0 -4.88 py + 1511 = 0

py = $ 309.63

px = $ 385.86

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