急 !急 ! 急 ! F4 Function 計法.

11a.
q = 12000 - 50(p-160)
11b.
Income, I = p q = p [12000 -50(p-160)]
I = 20000p - 50p^2
d(I)/dp = I' = 20000 - 100p
d(I')/dp = I'' = -100 < 0 => I will provide a maximum value when I' = 0
=> p = 200
thus, I = 2,000,000

you also can interpret the maximum value of I as below:
imagine y = Ax^2 + Bx + C, curve opens upwards when A < 0, such that it will gives a maximum value of y at x = -B/(2A)
I = -50p^2 + 20000p
p = -20000/(2 * -50) will gives maximum of I
p = 200 ... and I = 2,000,000

12a.
P = 2(2x+y) + 4x => 8x + 2y
A = 2x(2x+y) - 2x^2 = 2x^2 + 2xy
12b.
8x + 2y = 18 => y = 9 - 4x
A = 2x^2 + 2x(9-4x) => 18x - 6x^2
d(A)/dx = A' = 18 - 12x
d(A')/dx = A'' = -12 < 0 => A will provide a maximum value when A' = 0
=> x = 1.5
thus, A = 13.5 cm^2

2.
f(x) = 2x -1
=> f(y) = 2y - 1 or f(z) = 2z - 1
let y or z = x+1 => f(x+1) = 2(x+1) - 1 ...... OK