Maximum Product of Two Numbers whose sum is 30

This is a classic quadratic problem.
The two numbers are x and 30-x
Their product is (x)(30-x), which is a parabola.
The MAXIMUM of the parabola is the vertex.
You can find it in many ways, here, I use factoring.

[jakehobrath]

d/dx(30x-x^2)=30-2x, a quadratics have a min/max where dy/dx equals zero so 30-2x=0, x =15

[lechaiku]

This is just optimization problem that can be solved immediately by using the property of area optimization.

That's the reason why among all rectangles of a given perimeter, the square has the largest area.

It means that x = y

So when we have a such kind of problem (maximum product of two numbers) we need just one step.

If you have a sum (x+y) just divide it by 2.

30: 2 = 15 x = 15 and y = 15
That's it.

Or
when you don't have a sum but you have a product
A = xy

like in

P = x (30 -x)

what you need is only to make a comparison of one factor with other factor

x = 30 -x
2x = 30
x = 15

Also fast and easy.

[david]

Seems to me like you could also solve this with lagrange?

[BeaDSM]

But where's the proof? Also think it looks nicer to write it as P = (15 + x)(15 - x) as if you don't need to prove it is obvious then by inspection what value of x gives the maximum product (x = 0, so P(max) = 15 × 15 = 225).