APPLICATION OF QUADRATIC EQUATIONS The manager of a bicycle shop has found that at a price (in dollars ) of p(x)= 150 - x/5 per bicycle, x bicycles will be sold. a) Find an expression for the total revenue from the sale of x bicycles(hint: Revenue = demand x price) b)Find the number of bicycles sales that leads to maximum revenue c)find the maximum revenue. @timo86m
hmm i suppose the demand would be represented by x so 150x- x^2/5 as far as max and min take the derivative so 150-2/5x solve for x then plug back in for revenue! so b=375 c= 56250-28125= 28125
thanks but your solution is not really clear.if you can elaborate will be soo glad.thanks
where you got the 150-2/5x is where im confused
thats the derviative of 150x-(x^2)/5
so how do i go about solving for b
okay so by now you probably no that the derivative represents the slope correct?
yes
so when the slope =0 we assume its a max or min
since there is only one solution when i set the derivative to zero I assumed it was the max you can text for it through.
test*
so that x that we found is the number of bicycles being sold at that max
which was 375
I can explain this much easier on skype if you want
aww ma skype aint working now .the issue is the 375 NOW HOW YOU GOT THAT
okay so set the derivative to 0 and solve for x
BECAUSE OF THE X SQUARED I GET 750X - X^2=0
that is where i get to and then get confused
i have no idea how you end up with 750x-x^2
the derivative of 150x-(x^2)/5 is 150-2x/5
the first equation i just listed is the revenue
so how did you get 150-2x/5 how did u work to get that
n(x)(n-1)
n(x)^(n-1)*
so 150*1(x)^(1-1) n=1 in this one and 1/5*2x^(2-1) n=2
ok im tryna figure it out.will contact you if i have any question,.thanks.you soo helpful
let me know
the equation I listed needs to be put in vertex form
sorry not vertex form im dumb. standard form. which it is. -x^2+750x the axis of symmetry is what you are looking for
ax^2+bx+c so -b/2a, -750/2(-1)= 375
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