Minimize F = 3a + 4m if ma = 12. Assume a > 0, m > 0.

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.

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