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.
Answer:
It would cost $14.40 for six lollipops
Answer:
462 sq in
Step-by-step explanation:
3((14 2/3)(10.5))
3(154)
462
Answer:
a) 27 m/s
b) 30 m/s
c) i) 3
ii) Deceleration
Step-by-step explanation:
The question is not complete, the correct question is given as:
The graph shows information about the speed of a vehicle during the final 50 seconds of a journey. At the start of the 50 seconds the speed is k metres per second. The distance travelled during the 50 seconds is 1.35 kilometres.
(a) Work out the average speed of the vehicle during the 50 seconds
(b) Work out the value of k.
(c) (i) Calculate the gradient of the graph in the final 10 seconds of the journey
(ii) Describe what this gradient represents
Answer:
The graph is attached. The total time = 50 seconds, total distance = 1.35 km = 1350 m
a) The average speed is the ratio of the total distance traveled to the total time taken to cover this distance. The average speed is given by the formula:
b) From the graph, the total distance covered is the area of the graph. The graph is made up of a rectangle and triangle, the area of the graph is equal to the sum of area of rectangle and area of triangle.
c) i) The gradient in the last 10 seconds is the ratio of change in speed to change in time
ii) Since the gradient is negative it means it is deceleration. That is in the in the last 10 seconds the vehicle decelerates at a rate of 3 m/s²
Answer:
slope = 0
Step-by-step explanation:
y = - 
is the equation of a horizontal line parallel to the x- axis.
Since it is parallel to the x- axis the slope = 0
Thus any line parallel to it will have a slope = 0
