Answer:
Part A - see below
Part B :- Value of x for minimum cost = 20 meters
Step-by-step explanation:
I'll come back to you on Part A. For some reason I've got a mental block with this.
Part B ;- C(x) = 2x + 200 - √(x^2 - 300)
C(x) = 2x + 200 - (x^2 - 300)^1/2
Finding the derivative:-
C'(x) = 2 - 2x * 1/2(x^2 - 300)^ (-1/2)
C(x) = 2 - x(x^2 - 300)^-1/2
C(x) = 2 - x / (x^2-300)^1/2
C(x) = ( 2 (x^2 - 300)^1/2 - x ) / (x^2 - 300)^1/2
For minimum value this = 0 so
2 (x^2 - 300)^1/2 - x = 0
x^2 - 300 = (x/2)^2 (rearranging and squaring both sides)
x^2 - 300 = x^2/4
3x^2 / 4 = 300
x^2 = 300 * 4/3 = 400
x = 20
Part A
Lol!! I got muddled with this . I made it far to complicated than it is.
If the cost per meter of x is say 2 dollars the cost per meter for y = 1 dollars.
You can find y in terms of x by using Pythagoras:
x^2 = 300 + (200- y)^2
200 - y = √(x^2 - 300)
y = 200 - √(x^2 - 300)
so we have in terms of costs:-
cost of the new path in terms of x = 2*x + 1*(200 - √(x^2 - 300))
C(x) = 2x + 200 -√(x^2 - 300)
