maximization?

Question

I have to find maximization of f(a)=20(a-1)^1/2-2a-20...and I don't have a clue how to do it. I tried to differentiate it but my answer is wrong.

Answer 1

f(a) = 20(a-1)^(1/2) - 2a - 20
f'(a) = 10(a-1)^(-1/2) - 2
Set equal to zero and solve for a.
10(a-1)^(-1/2) - 2 = 0
Move (a-1)^(-1/2) to the denominator.
10/(a-1)^(1/2) - 2 = 0
Get a common denominator.
10/(a-1)^(1/2) - 2(a-1)^(1/2)/(a-1)^(1/2) = 0
[10 - 2(a-1)^(1/2)]/(a-1)^(1/2) = 0
Set the numerator equal to zero.
10 - 2(a-1)^(1/2) = 0
Subtract 10 from each side
-2(a-1)^(1/2) = -10
Divide both sides by -1
(a-1)^(1/2) = 5
Square each side.
a - 1 = 25
a = 26
Check to see if it is local max. (I will leave that for you)
a = 26 is indeed a local max.
f(26) = 20(26-1)^(1/2) - 2(26) - 20
f(26) = 20(5) - 52 - 20
f(26) = 100 - 72
f(26) = 28
Answer: (26,28) is a local max.

Answer 2

I assume you mean the maximum. First, you need to find the derivative of f(a). f'(a)=10(a-1)^(-1/2)-2. Let f'(a)=0 and solve for a. Thus 0=10(a-1)^(-1/2)-2 → 2=10(a-1)^(-1/2) → 1/5=(a-1)^(-1/2) → √(a-1)=5 → a-1=25 → a=26. Now just find the value of your original function at a=26 - this is f(26)=20√(26-1)-2(26)-20 →f(26)=20(5)-72 → f(26)=28. Thus your answer is (26, 28). It is easily verified (by evaluating f(25) and f(27), for instance) that this is indeed a maximum.

Answer 3

f(a) = 20(a-1)^1/2 - 2a - 20

f'(a) = (20/2)(a - 1)^(-1/2) - 2
= (10)/(a - 1)^(1/2) - 2

f'(a) = 0 (to find extrema)
0 = (10)/(a - 1)^(1/2) - 2
2 = (10)/(a - 1)^(1/2)
2(a - 1)^(1/2) = 10
(a - 1)^(1/2) = 5
a - 1 = 25
a = 26

If f''(26) < 0, then its a maximum:

f''(a) = (-5)(a - 1)^(-3/2)
f''(26) = (-5)(25)^3/2
= -625

x= 26 is a maximum, and
f(26) = 20(26 -1)^1/2 - 2(26) - 20 = 28

Maximization at (26,28)