calculus optimization problem?

Question

What is the product of the two largest positive numbers that have a sum of one(For example, 1/2 plus 1/2 is one and their product is 1/4. Is this the largest product possible? If so, what steps would you show to get this answer?)

Answer 1

x(1-x)=y
dy/dx=-x+(1-x)=0 for maximum
=>-2x+1=0=>x=1/2
d2y/dx^2=-2 negative so max.
x=1//2 and y=1/2

Answer 2

The product of the two largest numbers that add to one, will be the largest product as well, so maximize the PRODUCT:

P = xy

x + y = 1
y = 1 - x

P = x(1 - x)
=x - x^2

P' = 1 - 2x = 0
2x = 1
x = 1/2
and y = 1/2

Answer 3

Yes, be sure to check the second derivative to see if you have found a local max, local min, or inflection point.

However, this could have been done without calculus.

The equation to be maximized: x(1-x) becomes x-x^2

Hey, this is just a parabola and the extremum of a parabola is at the vertex. The vertex of a parabola is always at -b/2a when
y = ax^2 + bx + c. So, in this one, b = 1 and a = -1 so
the vertex is at x = -1/-2 = 1/2. But, you already knew that :)

Answer 4

Lets put the equation down as such:

x + y = 1

and we try to have x*y as large as possible.

For the equation we have

x = 1 - y

so we are trying to make (1-y) * y as large as possible.

(1-y)*y is y-y^2, and we know that the first derivative will be zero at a maximum.

The first derivative of y-y^2 is 1-2*y and if we put this as equal to zero, we have

1-2y=0
1=2y
1/2=y

With y equal to 1/2, we also have x equal to 1/2, since the sum of x and y is 1.

Answer 5

Yeah that's right. Think of the top half of a circle, when you get to the top of it the x,y are 1/2 and as you move along the circle in either direction the product of xy gets smaller.

Answer 6

x + y = 1 ==> y = 1 - x

P = xy = x - x^2

dP/dx = 1 - 2x = 0 ==> x = 1/2, y = 1/2

That's your answer.

Answer 7

let the two numbers be x and (1-x)
x(1-x)=x-x^2
when the product of two numbers is maximum,
d/dx(x-x^2)=0
1-2x=0
x=1/2

the two numbers should be both 1/2