Oh fuck I forgot the 2x with the chain :mad1:C = 5000 - 1000x + 2600(x^2 + 9)^1/2
dC/dx = - 1000 + 1300(x^2 + 9)^(-1/2) x 2x
= - 1000 + 2600x / (x^2 + 9)^(1/2)
min/max at dC/dx = 0
-1000 + 2600x / (x^2 + 9)^(1/2) = 0
-1 + 2.6x / (x^2 + 9)^(1/2) = 0
2.6x / (x^2 + 9)^(1/2) = 1 ---square both sides
6.76x^2 / (x^2 + 9) = 1
6.76x^2 = x^2 + 9
5.76x^2 = 9
x^2 = 25/16
x = 1.25 or -1.25 but x > 0
therefore x = 1.25
at x = 1, dC/dx = -ve
at x = 1.25 dC/dx = 0
at x = 1.5 dC/dx = +ve
therefore \_/ , minimum at x = 1.25
sub it in and the cost C = $12200
damnit, didnt check that, there goes 2 marksHe is right. Once you do that, you get x=root(300/7). But x is between 0 and 5. Hence it does not exist and the cheapest method is straight there.
You can't have SQ, SR because they are not connected to the power station then:okay i did none of what you guys are saying =.= i substitued my value for x (1.25 or something like that) into the lengths for SP and SQ and then found the costs for SP, SQ, SR and SR was the cheapest.. so yeah =.=
I don't think you can just say that. I differentiated and found a value that gives minimum again. But it was greater than 5, so I said since the doman of x is x must be less than or equal to 5, and x=5 cost in minimum.If you replaced 2.6 with 1.1, you will find that C = $7325
From P to S directly, the cost will be only root 34 x 1100 = $6414
From P to R , S to R , cost = 5(1000) + 3(1100) = $8300
Therefore cheapest route is from P to S
Most probably.Ok, well i got 1.25km ^^
But STUPID me, i forgot to calculate the total cost Zomg. !
is that 3/4 marks?
How many marks would I get for this? I basically done up to where the bold ends, but didn't have time (forgot I left this), so basically had it as a fraction not in simplest form. I then wrote to test, and that - 0 + = minimum... Also wrote "sub back into C" after it.C = 5000 - 1000x + 2600(x^2 + 9)^1/2
dC/dx = - 1000 + 1300(x^2 + 9)^(-1/2) x 2x
= - 1000 + 2600x / (x^2 + 9)^(1/2)
min/max at dC/dx = 0
-1000 + 2600x / (x^2 + 9)^(1/2) = 0
-1 + 2.6x / (x^2 + 9)^(1/2) = 0
2.6x / (x^2 + 9)^(1/2) = 1 ---square both sides
6.76x^2 / (x^2 + 9) = 1
6.76x^2 = x^2 + 9
5.76x^2 = 9
x^2 = 25/16
x = 1.25 or -1.25 but x > 0
therefore x = 1.25
at x = 1, dC/dx = -ve
at x = 1.25 dC/dx = 0
at x = 1.5 dC/dx = +ve
therefore \_/ , minimum at x = 1.25
sub it in and the cost C = $12200
Most probably: 1 mark for differentation, 1 mark for finding x, 1 mark for testing, 1 mark for substituting back to find the cost. So maybe 1-2 for you.How many marks would I get for this? I basically done up to where the bold ends, but didn't have time (forgot I left this), so basically had it as a fraction not in simplest form. I then wrote to test, and that - 0 + = minimum... Also wrote "sub back into C" after it.
Would I get 3 or 2? Obviously I didn't have time to get the next 2 marks either...
They pretty much give you the equation C=1000 (5-x+2.6 sqrt(x^2+9)) and I assumed you have to find the first derivative of it to get the stationary points and ultimately the minimum, but when I tried to differentiate and solve for dC/dx = 0 I ended up with x^2+9=13/5, which does not have any real solutions! So what happens now? Did I do something wrong??