Suppose a snowboard manufacturer will sell q snowboards per week when charging p dollars per snowboard where q= (600−p)^3. Find the price the company should charge to maximize revenue? (Recall:R=pq, revenue equals quantity times price.)

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Answer 1 · 2017-12-07T07:36:05

The company should charge #$ 150# per snowboard to get

the maximum revenue of #$ 13668750000# by way of selling

#91125000# snowboards per week.

#"The Revenue "R=pq=p(600-p)^3.#

This is a function of #p,# so let us write,

# R(p)=p(600-p)^3.#

For #R_max, R'(p)=0, and, R''(p) lt 0.#

Now, #R(p)=p(600-p)^3.#

Using the Product Rule, to differentiate, we get,

# R'(p)=pd/(dp){(600-p)^3}+(600-p)^3d/(dp){p},#

#=p*3(600-p)^2*d/(dp){(600-p)}+(600-p)^3*1.#

# =-3p(600-p)^2+(600-p)^3.#

#=(600-p)^2{(600-p)-3p},#

#rArr R'(p)=4(600-p)^2(150-p).#

Also, #R''(p),#

#=4{(600-p)^2*d/(dp)(150-p)+(150-p)*d/(dp)(600-p)^2},#

#=4{(600-p)^2*(-1)+(150-p)(2(600-p))(-1)},#

#=4{-(600-p)^2-2(150-p)(600-p)},#

#=-4(600-p){600-p+2(150-p)},#

#=-4(600-p)(900-3p),#

#rArr R''(p)=-12(600-p)(300-p).#

Now, #R'(p)=0 rArr p=600, or, p=150.#

Also, #R''(600)=0, & R''(150)=-12*450*150 lt 0.#

Therefore, #p=150" gives "R_max=R(150)=13668750000$.#

Thus, the company should charge #$ 150# per snowboard to get

the maximum revenue of #$ 13668750000# by way of selling

#91125000# snowboards per week.

Enjoy Maths.!