Solve Min Cost of Fencing Rectangular Land w/ 2 Eqns

[Chocolaty]

Ok we're doing optimization, i don't understand why you have to derive, it just doesn't come naturally to me, like it's not something i instinctively think of doing...

My example

A company wants to fence a rectangular piece of land that is bordered on one side by a road and by a river on the opposite side. There will be no fence along the river. The fence along the road costs $3 per foot and the fence along the other two sites costs $2 per foot. If the rectangular piece of land must have an area of 10800 square feet, find the dimensions that will give the minimum cost. What is the minimum cost?

I understand the need to use 2 equations like in related rates to help reduce, in this case, the number of variables in the cost function.

Eq#1: Min C = 3x + 4y
Eq#2: xy = 10800
y = 10800/x
Min C = 3x + 4(10800/x)
Min C = 3x + 43200/x [x > 0]

At this point... he derives, but why?
C' = 3 - 43200/(x^2)

Now he finds the critical numbers, why? And also, how can you just take the right side of the equation of C' to do that?
CN: 3 = 43200/(x^2)
3x^2 = 43200
x^2 = 14400
x = 120

Now he finds the second derivative, again, why?
C'' = 86400/(x^3)
C(120) > 0 => x = 120 will give Min C

Ans: x = 120, y = 90, Min C = $720

 

[Chocolaty]

Nevermind guys i found out. The reason why you need to find the first derivative is not "because it's fun" but because it's the only way to find the critical numbers (which are candidates for rel. maximum and rel. minimum). So you find your first derivative then you find the critical numbers, which in this cases is x = 120. After that you find the second derivative to find if this point (where x = 120) is located in a concave or concave down part of the graph. If when you find the second derivative of this point it's > 0 then it's a minimum for sure. If it's < 0 then it's a maximum for sure.