Find the maximum value of the product of two real numbers

[chwala]

Homework Statement

The sum of two real numbers $x$ and $y$ is $12$. Find the maximum value of their product $xy$.

Relevant Equations

Arithmetic and geometric means

Using the inequality of arithmetic and geometric means,
$$\frac {x+y}{2}≥\sqrt{xy}$$
$$6^2≥xy$$
$$36≥xy$$

I can see the textbook answer is $36$, my question is can $x=y?$, like in this case.

 

[anuttarasammyak]

Solve
xy=36
x+y=12
to check your assumption.

 

[chwala]

Solve
xy=36
x+y=12
to check your assumption.

$$x^2-12x+36=0$$, Giving us a repeated root... i get that, ok $⇒x=y$ cheers Anutta...

 

[Mark44]

Another way to do this that doesn't use the arithmetic mean and geometric mean:

Maximize $f(x, y) = xy$ given that $x + y = 12$.
$f(x, y) = xy = x(12 - x) = -x^2 + 12x = -(x^2 - 12x + 36) + 36 = -(x - 6)^2 + 36$
The graph of the last expression is a parabola that opens downward, with its vertex at (6, 36). Since the parabola opens downward, its maximum value is 36.

 

[WWGD]

Yet another way is direct substitution: use $x+y =6$ to sub in in $f(x,y)= xy$ and get a function of x alone. Then find the max in the usual way.

 

[Mark44]

Yet another way is direct substitution: use $x+y =6$ to sub in in $f(x,y)= xy$ and get a function of x alone. Then find the max in the usual way.

This is exactly what I did in my previous post.

 

[vela]

In both $x+y=12$ and $xy$, $x$ and $y$ appear symmetrically, so I'd expect $x=y$ to correspond to an extremum $xy$. It's then easy to see that it's a maximum and not a minimum.

 

[WWGD]

In both $x+y=12$ and $xy$, $x$ and $y$ appear symmetrically, so I'd expect $x=y$ to correspond to an extremum $xy$. It's then easy to see that it's a maximum and not a minimum.

Good point. Guess we can model it by a rectangle with perimeter 24 ( simplified by halving), whose area is maximized when both its sides are equal.

 

[chwala]

...I just noted that we could also use the Lagrange Multiplier, in solving this kind of problems...

 

[benorin]

Lagrange multipliers method:

$f(x,y)=xy$
$g(x,y)=x+y=12$

$\nabla f := \lambda \nabla g$
$<y,x> := \lambda <1,1>$
Hence $x=\lambda =y\Rightarrow x+x=12\Rightarrow x=6=y$

Not so bad :)

 

[chwala]

Lagrange multipliers method:

$f(x,y)=xy$
$g(x,y)=x+y=12$

$\nabla f := \lambda \nabla g$
$<y,x> := \lambda <1,1>$
Hence $x=\lambda =y\Rightarrow x+x=12\Rightarrow x=6=y$

Not so bad :)

Nice,...$g(x,y)= x+y$...from what I've read... and not $x+y-12$...Is that correct. Cheers man!

 

[benorin]

the derivative of a constant is zero so it doesn't matter

 

[chwala]

the derivative of a constant is zero so it doesn't matter

Thanks ...I had already noted that...I just want to be certain on the correct definition of $g(x,y)$...cheers benorin.

 

[benorin]

Don't think of it as a function $g(x,y)$ think of it as a restriction $g(x,y):=k$ is an equation not a function, more precisely it's level curve or surface of the function I told you not to think of. ;)