Hey, can anyone me with this optimisation question? tia x

4

Step-by-step explanation:

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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)

answer:  

123455421

step-by-step explanation: easy