# Find two numbers whose sum is 23 and whose product is a maximum [Solved] x y=23
z=xy=y(23-y)=23y-y^2
dz/dy=23-2y=0 to be max
y=11.5
x=11.5

Product : 529/4

Im French, sorry for language.

1 22=23 1*23=22
2 21=23 ..2*23=42
3 20=23..2*23=60
so on
10 13=23 ..10*13=130
11 12=23.11*23=132 [This is your max product] so the answer is 11 and 12
13 10=23.13*23=130
After the max product has been reached the numbers will start going down and repeating it self.

xy = x(23 x) = -x^2 23x which we know is a frowny face parabola =). But this is good because we know that this type of parabola will have a maximum value! What is the maximum value Or more specifically where does the value occur It will occur at the vertex!

x y = 23

So we will let the two numbers be x and y.

x y = 23

Using calculus, we take the derivative of -x^2 23x to get -2x 23 and set this equal to 0 to find where the graph has a horizontal tangent. It will occur at -2x 23=0 => -2x = -23 => x=23/2

1 22=23 1*23=22
2 21=23 ..2*23=42
3 20=23..2*23=60
so on
10 13=23 ..10*13=130
11 12=23.11*23=132 [This is your max product] so the answer is 11 and 12
13 10=23.13*23=130
After the max product has been reached the numbers will start going down and repeating it self.

What is the value of the parabola at x = 23/2 its just -(23/2)^2 23(23/2) = -132.25 264.5 = 132.25

So we know that that the maximum value of the product will be 132.25 and the two numbers whose sum is 23 will be x = 23/2 and y = 23 (23/2) = 23/2

Good luck!

You could do this with calculus, but the quick way is to set them equal. 11.5*11.5 = 132.25

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x y = 23

y = 23/2

y = 23/2

Product : 529/4

You could do this with calculus, but the quick way is to set them equal. 11.5*11.5 = 132.25

P = 529/4 529/2

y = 23/2

Substituting this into our expression xy we get:

11,12

x y = 23

If you mean whole numbers, it would be 11 and 12.

Good luck!

What is the value of the parabola at x = 23/2 its just -(23/2)^2 23(23/2) = -132.25 264.5 = 132.25

We want to maximize their product xy. But how do we do this without having a function of one variable

Polynomial like : ax bx c, where :
a = 1
b = 23
c = 529/4

Im French, sorry for language.

Im French, sorry for language.

Using calculus, we take the derivative of -x^2 23x to get -2x 23 and set this equal to 0 to find where the graph has a horizontal tangent. It will occur at -2x 23=0 => -2x = -23 => x=23/2

I gues 11 and 12

Polynomial like : ax bx c, where :
a = 1
b = 23
c = 529/4

Using calculus, we take the derivative of -x^2 23x to get -2x 23 and set this equal to 0 to find where the graph has a horizontal tangent. It will occur at -2x 23=0 => -2x = -23 => x=23/2

11,12

11.5 and 11.5

Product : 529/4

Using calculus, we take the derivative of -x^2 23x to get -2x 23 and set this equal to 0 to find where the graph has a horizontal tangent. It will occur at -2x 23=0 => -2x = -23 => x=23/2

1 22=23 1*23=22
2 21=23 ..2*23=42
3 20=23..2*23=60
so on
10 13=23 ..10*13=130
11 12=23.11*23=132 [This is your max product] so the answer is 11 and 12
13 10=23.13*23=130
After the max product has been reached the numbers will start going down and repeating it self.

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Substituting this into our expression xy we get:

xy = x(23 x) = -x^2 23x which we know is a frowny face parabola =). But this is good because we know that this type of parabola will have a maximum value! What is the maximum value Or more specifically where does the value occur It will occur at the vertex!

I gues 11 and 12

P = 529/4

So we will let the two numbers be x and y.

If you mean whole numbers, it would be 11 and 12.

x(23 x) = 529/4

y = 23 23/2

23x x = 529/4

1 22=23 1*23=22
2 21=23 ..2*23=42
3 20=23..2*23=60
so on
10 13=23 ..10*13=130
11 12=23.11*23=132 [This is your max product] so the answer is 11 and 12
13 10=23.13*23=130
After the max product has been reached the numbers will start going down and repeating it self.

1 22=23 1*23=22
2 21=23 ..2*23=42
3 20=23..2*23=60
so on
10 13=23 ..10*13=130
11 12=23.11*23=132 [This is your max product] so the answer is 11 and 12
13 10=23.13*23=130
After the max product has been reached the numbers will start going down and repeating it self.

We want to maximize their product xy. But how do we do this without having a function of one variable

= (- 23) 4(1 * 529/4) = 23 529 = 529 529 = 0

Well we can solve our first equation for y to get y = 23 x.

Hen, you can write with the 2 numbers :