Proving the highest product of two numbers with a fixed sum
$begingroup$
How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?
Example:
Given sum equals $10$.
Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?
products
edited Jan 15 at 18:24
David G. Stork
12.2k41836
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
$endgroup$
add a comment |
$begingroup$
How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?
Example:
Given sum equals $10$.
Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?
products
edited Jan 15 at 18:24
David G. Stork
12.2k41836
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
$endgroup$
3
$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35
$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36
$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39
add a comment |
$begingroup$
How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?
Example:
Given sum equals $10$.
Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?
products
edited Jan 15 at 18:24
David G. Stork
12.2k41836
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
$endgroup$
How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?
Example:
Given sum equals $10$.
Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?
products
products
edited Jan 15 at 18:24
David G. Stork
12.2k41836
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
edited Jan 15 at 18:24
David G. Stork
12.2k41836
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
edited Jan 15 at 18:24
David G. Stork
12.2k41836
edited Jan 15 at 18:24
David G. Stork
12.2k41836
edited Jan 15 at 18:24
David G. Stork
12.2k41836
12.2k41836
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
asked Jan 15 at 17:32
Xxx DddXxx Ddd
554
554
3
$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35
$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36
$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39
add a comment |
3
$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35
$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36
$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39
3
3
$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35
$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35
$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36
$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36
$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39
$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
$4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
$endgroup$
$begingroup$
Mark.In 1 line! Very nice.
$endgroup$
– Peter Szilas
Jan 15 at 18:10
add a comment |
$begingroup$
Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
add a comment |
$begingroup$
Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
add a comment |
$begingroup$
You want two numbers which give a particular sum and you want to maximize their product:
$$m+n = c tag{1}$$
$$m(n) = ? tag{2}$$
You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get
$$m(c-m) = -m^2+cm$$
If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).
answered Jan 15 at 17:53
KM101KM101
6,0861525
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words.
Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
add a comment |
$begingroup$
$$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
$endgroup$
$begingroup$
Mark.In 1 line! Very nice.
$endgroup$
– Peter Szilas
Jan 15 at 18:10
add a comment |
$begingroup$
$4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
$endgroup$
$begingroup$
Mark.In 1 line! Very nice.
$endgroup$
– Peter Szilas
Jan 15 at 18:10
add a comment |
$begingroup$
$4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
$endgroup$
$4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
answered Jan 15 at 17:53
Mark BennetMark Bennet
82.1k984182
82.1k984182
$begingroup$
Mark.In 1 line! Very nice.
$endgroup$
– Peter Szilas
Jan 15 at 18:10
add a comment |
$begingroup$
Mark.In 1 line! Very nice.
$endgroup$
– Peter Szilas
Jan 15 at 18:10
$begingroup$
Mark.In 1 line! Very nice.
$endgroup$
– Peter Szilas
Jan 15 at 18:10
$begingroup$
Mark.In 1 line! Very nice.
$endgroup$
– Peter Szilas
Jan 15 at 18:10
add a comment |
$begingroup$
Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
add a comment |
$begingroup$
Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
add a comment |
$begingroup$
Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Jan 15 at 17:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
79.6k42867
add a comment |
add a comment |
$begingroup$
Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
add a comment |
$begingroup$
Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
add a comment |
$begingroup$
Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
answered Jan 15 at 17:38
Thomas ShelbyThomas Shelby
4,8082727
4,8082727
add a comment |
add a comment |
$begingroup$
You want two numbers which give a particular sum and you want to maximize their product:
$$m+n = c tag{1}$$
$$m(n) = ? tag{2}$$
You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get
$$m(c-m) = -m^2+cm$$
If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).
answered Jan 15 at 17:53
KM101KM101
6,0861525
$endgroup$
add a comment |
$begingroup$
You want two numbers which give a particular sum and you want to maximize their product:
$$m+n = c tag{1}$$
$$m(n) = ? tag{2}$$
You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get
$$m(c-m) = -m^2+cm$$
If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).
answered Jan 15 at 17:53
KM101KM101
6,0861525
$endgroup$
add a comment |
$begingroup$
You want two numbers which give a particular sum and you want to maximize their product:
$$m+n = c tag{1}$$
$$m(n) = ? tag{2}$$
You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get
$$m(c-m) = -m^2+cm$$
If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).
answered Jan 15 at 17:53
KM101KM101
6,0861525
$endgroup$
You want two numbers which give a particular sum and you want to maximize their product:
$$m+n = c tag{1}$$
$$m(n) = ? tag{2}$$
You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get
$$m(c-m) = -m^2+cm$$
If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).
answered Jan 15 at 17:53
KM101KM101
6,0861525
answered Jan 15 at 17:53
KM101KM101
6,0861525
answered Jan 15 at 17:53
KM101KM101
6,0861525
answered Jan 15 at 17:53
KM101KM101
6,0861525
6,0861525
add a comment |
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words.
Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words.
Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words.
Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
Sometimes a figure is worth 1000 words.
Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
edited Jan 15 at 18:14
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
answered Jan 15 at 17:44
David G. StorkDavid G. Stork
12.2k41836
12.2k41836
add a comment |
add a comment |
$begingroup$
$$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
$endgroup$
add a comment |
$begingroup$
$$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
$endgroup$
add a comment |
$begingroup$
$$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
$endgroup$
$$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
answered Jan 15 at 18:56
Will JagyWill Jagy
105k5103202
105k5103202
add a comment |
add a comment |
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$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35
$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36
$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39