Ashley opened a lemonade stand on her street. It costs her $12.50 to buy the supplies each day, and she sells lemonade for $2.50
1 answer:
You might be interested in
=[(sinx/cosx)/(1+1/cosx)] + [(1+1/cosx)/(sinx/cosx)]
=[(sinx/cosx)/(cosx+1/cosx)]+[(cosx+1/cosx)/(sinx/cosx)]
= [sinx/(cosx+1)] + [(cosx+1)/sinx]
= [sin^2x+(cosx+1)^2] / [sinx (cosx+1)]
= [2+2cosx] / [sinx(cosx+1)]
=[2(cosx+1)] / [sinx (cosx+1)]
= 2/sinx
= 2 cscx
(I think this will be helpful for you. if you can see the picture, it has more detail in it.)
=[(sinx/cosx)/(cosx+1/cosx)]+[(cosx+1/cosx)/(sinx/cosx)]
= [sinx/(cosx+1)] + [(cosx+1)/sinx]
= [sin^2x+(cosx+1)^2] / [sinx (cosx+1)]
= [2+2cosx] / [sinx(cosx+1)]
=[2(cosx+1)] / [sinx (cosx+1)]
= 2/sinx
= 2 cscx
(I think this will be helpful for you. if you can see the picture, it has more detail in it.)
Q(p) = k/p^3 . . . . . . . . we want to find k
q(10) = k/10^3 = 64
k = 64,000
Revenue = q(p)*p = 64000/p^2
Cost = 150 +2q = 150 +2*64000/p^3
Profit = Revenue -Cost = 64000/p^2(1 -2/p) -150
Differentiating to find the maximum profit, we have
.. dProfit/dp = -2(64000/p^3) +6(64000/p^4) = 0
.. -1 +3/p = 0
.. p = 3
A price of $3 per unit will yield a maximum profit.
q(10) = k/10^3 = 64
k = 64,000
Revenue = q(p)*p = 64000/p^2
Cost = 150 +2q = 150 +2*64000/p^3
Profit = Revenue -Cost = 64000/p^2(1 -2/p) -150
Differentiating to find the maximum profit, we have
.. dProfit/dp = -2(64000/p^3) +6(64000/p^4) = 0
.. -1 +3/p = 0
.. p = 3
A price of $3 per unit will yield a maximum profit.
Answer:
honestly,I also really didn't know the ans so I took it from internet. hope you won't mind
