The answer is J'(6, -2) M'(5,1)
2xy + 5x -12y -30
x(2y + 5) - 6( 2y + 5)
(2y+5) (x-6)
Answer:
C(p) = 4,96 (in thousands of dollars)
l = 2980 $ invest in labor
k = 2980 $ invest in equipment
Step-by-step explanation:
Information we have:
Monthly output P = 450*l*k ⇒ k = P/450*l
But the production need to be 4000
Then k = 4000/450*l
Cost of production = l * k (in thousands of dollars)
C(l) = l + 4000/450*l
Taking derivatives (both members of the equation)
C´(l) = 1 - 400 /45*l² ⇒ C´(l) = 0 ⇒ 1 - 400/45l² = 0
45*l² - 400 = 0 ⇒ l² = 400/45
l = 2.98 (in thousands of dollars)
l = 2980 $ And
k = 400/45*l ⇒ k 400/45*2.98
k = 2.98 (in thousands of dollars)
C(p) = l + k
C(p) = 2980 + 2980
C(p) = 5960 $
It’s funny cause I have a friend named Rashawn
But y intercept would be 550 then multiply 8 by 6to get 48
So your answer should be $598 in 6 years
