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FinnZ [79.3K]
2 years ago
11

Jim's Camera shop sells two high-end cameras, the Sky Eagle and Horizon. The demands 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 = 229 − 0.60PS + 0.35PH
DH = 269 + 0.10PS − 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. (Enter your answer in terms of
PS and PH.)
revenue =
Mathematics
1 answer:
BartSMP [9]2 years ago
5 0

The revenue is given by the product of the price and the demand function

  • The revenue function for Sky Eagle is <u>RS = 229·PS - 0.60·PS² + 0.35·PS·PH</u>.
  • The revenue function for Horizon is <u>RH = 269·PH + 0.10·PS·PH - 0.64·PH²</u>.

Reasons:

The given parameters are;

The demand function for the Sky Eagle, DS = 229 - 0.60·PS + 0.35·PH

The demand function for the Horizon, DH = 269 + 0.10·PS - 0.64·PH

The above demand functions gives the quantity demanded

Therefore, the revenue function is given by the product of the price and the demand function as follows;

RS = PS × DS = PS × (229 - 0.60·PS + 0.35·PH) = 229·PS - 0.60·PS² + 0.35·PS·PH

The revenue for Sky Eagle <u>RS = 229·PS - 0.60·PS² + 0.35·PS·PH</u>

<u />

Revenue for Horizon is found as follows;

RH = PH ×(269 + 0.10·PS - 0.64·PH) = 269·PH + 0.10·PS·PH - 0.64·PH²

The revenue for Horizon, <u>RH =  269·PH + 0.10·PS·PH - 0.64·PH²</u>

Learn more demand function here:

brainly.com/question/16888880

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You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3
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Answer:

Dimensions will be

Length = 7.23 cm

Width = 7.23 cm

Height = 9.64 cm

Step-by-step explanation:

A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

h=\frac{504}{lw}

Sides which involve length and width and height, cost = 3 cents per cm²

Top and bottom of the box costs = 4 cents per cm²

Cost of the sides C_{s}= 3[2(l + w)h] = 6(l + w)h

C_{s}= 3[2(l + w)h]

C_{s}=6(l+w)(\frac{504}{lw} )

Cost of the top and the bottom C_{(t,p)}= 4(2lw) = 8lw

Total cost of the box C = 3024\frac{(l+w)}{lw} + 8lw

                                      = 3024[\frac{1}{l}+\frac{1}{w}] + 8lw

To minimize the cost of the sides

\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0

\frac{3024}{l^{2}}=8w

\frac{378}{l^{2}}=w ---------(1)

\frac{dC}{dw}=3024(-w^{-2})+8l=0

\frac{3024}{w^{2}}=8l

\frac{378}{w^{2}}=l

w^{2}=\frac{378}{l}-------(2)

Now place the value of w from equation (1) to equation (2)

(\frac{378}{l^{2}})^{2}=\frac{378}{l}

\frac{(378)^{2} }{l^{4}}=\frac{378}{l}

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

w^{2}=\frac{378}{7.23}

w^{2}=52.28

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

h=\frac{504}{(7.23)^{2}}

h = 9.64 cm                        

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