A smokestack deposits soot on the ground with a concentration inversely proportional to the square of the distance from the stack. With two smokestacks 20 miles apart, the concentration of the combined deposits on the line joining them, at a distance x from one stack, is given by

S = ksub1 over x^2 + ksub2/((20-x)^2)
where ksub1 and ksub 2
are positive constants which depend on the quantity of smoke each stack is emitting. If
ksub1 = 2ksub2,
find the point on the line joining the stacks where the concentration of the deposit is a minimum. (Round your answer to two decimal places.)

so:
s= 11k2/x^2 + k2/((20-x)^2) = 11k2((20-x)^2) + x^2k2 all divided by x^2((20-x)^2)
so then I multiplied it out:

11k2(x^2-40x+400)+x^2k2
11k2x^2-440k2x+4400k2+x^2k2

and then I didn't know where to go from there unfortunately

Thank you so much for helping me understand how to solve the problem!

To find the point on the line where the concentration of deposit is a minimum, we need to find the minimum value of the function S(x).

To do this, we can take the derivative of S(x) with respect to x, and set it equal to zero to find the critical points. Then we can evaluate the second derivative to determine whether the critical point is a minimum or maximum.

Let's find the derivative of S(x) with respect to x:

dS/dx = d/dx [11k2x^2 - 440k2x + 4400k2 + x^2k2]
= 22k2x - 440k2 - 2xk2

Setting this equal to zero, we get:

22k2x - 440k2 - 2xk2 = 0

Simplifying,

20k2x - 440k2 = 0
20k2x = 440k2
x = 22

So, the critical point is x = 22.

Now, let's evaluate the second derivative to determine whether it is a minimum or maximum:

d2S/dx2 = d/dx [22k2x - 440k2 - 2xk2]
= 22k2 - 2k2
= 20k2

Since k2 is a positive constant, the second derivative is positive. Therefore, the critical point x = 22 corresponds to a minimum.

Thus, the point on the line joining the stacks where the concentration of the deposit is a minimum is at x = 22 miles from one stack.

To find the point on the line joining the stacks where the concentration of the deposit is a minimum, we need to find the value of x that minimizes the expression S.

Let's start by simplifying the expression you have:

S = (11k2(x^2-40x+400) + x^2k2) / (x^2((20-x)^2))

Expanding the numerator, we get:

S = (11k2x^2 - 440k2x + 4400k2 + x^2k2) / (x^2((20-x)^2))

Combining like terms, we have:

S = ((11k2 + k2)x^2 - 440k2x + 4400k2) / (x^2((20-x)^2))

Now, let's set up the equation for finding the minimum:

To find the minimum, we need to take the derivative of S with respect to x, set it equal to zero, and solve for x.

dS/dx = 0

To simplify the equation, let's multiply through by (x^2)((20-x)^2):

((11k2 + k2)x^2 - 440k2x + 4400k2) = 0

Expand and rearrange:

12k2x^2 - 440k2x + 4400k2 = 0

Now, we have a quadratic equation. We can solve it using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 12k2, b = -440k2, and c = 4400k2.

Plugging in the values, we get:

x = (-(-440k2) ± √((-440k2)^2 - 4(12k2)(4400k2))) / (2(12k2))

Simplifying further:

x = (440k2 ± √((440k2)^2 - 4(12k2)(4400k2))) / (24k2)

x = (440k2 ± √(193600k4 - 21120k4)) / (24k2)

x = (440k2 ± √(172480k4)) / (24k2)

x = (440k2 ± 416k2) / (24k2)

Now, we can simplify it further:

x = (440 ± 416) / 24

There are two possible solutions:

x1 = (440 + 416) / 24 = 856 / 24 = 35.67

x2 = (440 - 416) / 24 = 24 / 24 = 1

Since the distance between the two smokestacks is 20 miles, the value of x must lie between 0 and 20.

Therefore, the point on the line joining the stacks where the concentration of the deposit is a minimum is at x = 1 mile.

Let's call the k's a and b just to make things more readable.

S = a/x^2 + b/(20-x)^2
If a = 2b, then
S = 2b/x^2 + b/(20-x)^2
Now, we can actually ignore b, since it is just a constant multiplier and will not affect where the minimum occurs. So, let's use

S = 2/x^2 + 1/(20-x)^2
Now we have
dS/sx = -4/x^3 + 2/(20-x)^3
=

-2(3x^3-120x^2+2400x-16000)
--------------------------------
x^3 (20-x)^3

All we need to examine is the numerator.

dS/dx=0 at x ≈ 11.15