Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
Answer:
Option D is correct ...
Step-by-step explanation:
because f(x) is defined on x<0 which is only possible in log(-x)
Answer:
The price where the manufacture sells the maximum number of toys is $20
Step-by-step explanation:
The given equation for that represents the number of toys the manufacturer can sell is given as follows;
T = -4·p² + 160·p - 305
Where;
p = The price of the toys in dollars
At the point where the manufacture sells the maxim number of toys on the graph of the equation T = -4·p² + 160·p - 305, which is the top of the graph, the slope = 0
Therefore, at the maximum point;
The slope = 0 = dT/dp = d(-4·p² + 160·p - 305)/dp = -8·p + 160
∴ -8·p + 160 = 0
160 = 8·p
8·p = 160
p = 160/8 = 20
The price where the manufacture sells the maximum number of toys is = p = 20 dollars
