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.
Area is base x height
12 x 6 = 72
the area is 72 in^2
1/(x^5) is equal to x^-5, by the definition of negative exponents.
<span>100 = 0.046s^2 - 0.199s + 0.264
0 = 0.046s^2 - 0.199s - 99.736
s= 48.77691127... or -44.45082431 (reject)
The car's speed was approximately 48.8 mph.</span>