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)

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.