Min value of F = 24
It happens when m = 3 and a = 4.
Explanation:
Start with ma = 12 and Solve for 'a' to get a = 12/m
Plug it into the other equation to get
F = 3a + 4m
F = 3(12/m) + 4m
F = 36/m + 4m
F = 36/m + (4m^2)/m
F = (36+4m^2)/m
The equation above is in the form F = a/b where a = 36+4m^2 and b = m
Applying derivatives to each piece gives a' = 8m and b' = 1
So,
F = a/b
F ' = (a/b)'
F ' = (a' * b - a * b')/(b^2) quotient rule
F ' ( 8m*m - (36+4m^2)*1 )/(m^2)
F ' = (8m^2 - 36 - 4m^2)/(m^2)
F ' = (4m^2 - 36)/(m^2)
The minimum of F occurs when F ' = 0
F ' = 0
(4m^2 - 36)/(m^2) = 0
4m^2 - 36 = 0*m^2
4m^2 - 36 = 0
4m^2 = 36
m^2 = 36/4
m^2 = 9
m = sqrt(9) or m = -sqrt(9)
m = 3 or m = -3
Keep in mind that m > 0, so we ignore m = -3
Next we'll use the first derivative test. I find it's easier compared to the second derivative test because we don't have to do another derivative.
The function F(m) = (36+4m^2)/m has a potential min point at m = 3.
To see if we have an actual min point or not, we need to check the sign of F ' (m) when m = 2 and m = 4; that way we see if F ' (m) changes sign as we pass through m = 3.
First compute the derivative when m = 2
F ' (m) = (4m^2 - 36)/(m^2)
F ' (2) = (4(2)^2 - 36)/(2^2)
F ' (2) = (4*4-36)/(4)
F ' (2) = (16-36)/(4)
F ' (2) = -20/4
F ' (2) = -5
The actual value doesn't matter. All we're after is the sign of it. So we see that F ' (2) is negative which means we know that F(m) is decreasing when 0 < m < 3
Now let's try m = 4
F ' (m) = (4m^2 - 36)/(m^2)
F ' (4) = (4(4)^2 - 36)/(4^2)
F ' (4) = (4*16-36)/(16)
F ' (4) = (64-36)/(16)
F ' (4) = 28/4
F ' (4) = 7
This value is positive, so F(m) is increasing on the interval 0 < m < infinity
F(m) decreases on 0 < m < 3 and increases on 3 < m < infinity
So the F function goes downhill and then goes back uphill, and this lowest valley point is when m = 3
So we've confirmed that F(m) does indeed have a min value at m = 3.
Making a sign chart or interval might help visualize it.
Now onto the last part.
Plug m = 3 into F(m) to find the min value of F
F = (36+4m^2)/m
F = (36+4*3^2)/3
F = (36 + 4*9)/3
F = (36 + 36)/3
F = 72/3
F = 24
The smallest F can get is 24 and it happens when m = 3 and a = 12/m = 12/3 = 4.
Side note: A non-calculus approach would have you graphing y = (36+4x^2)/x, and then using the graphing calculator to find the lowest point in the first quadrant. We're in this quadrant since a, m and F are all positive.
