Mixing
Exercise XV num 6 from Silvanus Thompson - Calculus Made Easy
0
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Exercise asks to verify that the sum of three quantities x, y, z, whose product is a constant k, is minimum when these three quantities are equal.
This is my amateurish attempt:
$x + y + z = S$;
$x*y*z=k$;
$z = frac{k}{xy}$;- x + y + $frac{k}{xy}$ = S;
$frac{partial S}{partial x} = 1 - frac{k}{x^2y} = frac{x^2y-k}{x^2y}$ and symmetricly $frac{partial S}{partial y} = frac{y^2x-k}{y^2x}$;
$x^2y-k = 0 rightarrow x^2y=k rightarrow x^2y=xyz rightarrow frac{x^2y}{xy}=z rightarrow x=z$ and symmetricly $y=z$, so $y=x=z$;- This part is doubtful to me. Proving that S is minimal when $x=y=z rightarrow$ $frac{partial ^2 S}{partial x^2} = frac{2k}{yx^3} rightarrow frac{2(xyz)}{yx^3} rightarrow frac{2z}{x^2} rightarrow frac{2x}{x^2} rightarrow frac{2}{x}$ and symmetricly $frac{2}{y}$;
Conclusion - if second derivative is positive, that means $S$ must be a minimum.
Please be free to comment if this solution is flawed or missing something.
calculus optimization
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
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|
show 1 more comment
0
$begingroup$
Exercise asks to verify that the sum of three quantities x, y, z, whose product is a constant k, is minimum when these three quantities are equal.
This is my amateurish attempt:
$x + y + z = S$;
$x*y*z=k$;
$z = frac{k}{xy}$;- x + y + $frac{k}{xy}$ = S;
$frac{partial S}{partial x} = 1 - frac{k}{x^2y} = frac{x^2y-k}{x^2y}$ and symmetricly $frac{partial S}{partial y} = frac{y^2x-k}{y^2x}$;
$x^2y-k = 0 rightarrow x^2y=k rightarrow x^2y=xyz rightarrow frac{x^2y}{xy}=z rightarrow x=z$ and symmetricly $y=z$, so $y=x=z$;- This part is doubtful to me. Proving that S is minimal when $x=y=z rightarrow$ $frac{partial ^2 S}{partial x^2} = frac{2k}{yx^3} rightarrow frac{2(xyz)}{yx^3} rightarrow frac{2z}{x^2} rightarrow frac{2x}{x^2} rightarrow frac{2}{x}$ and symmetricly $frac{2}{y}$;
Conclusion - if second derivative is positive, that means $S$ must be a minimum.
Please be free to comment if this solution is flawed or missing something.
calculus optimization
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
$endgroup$
$begingroup$
Did the book mention the Inequality of arithmetic and geometric means?
$endgroup$
– A.Γ.
Jan 2 at 13:18
$begingroup$
Were any conditions placed on k?
$endgroup$
– William Elliot
Jan 2 at 13:44
$begingroup$
@WilliamElliot k is just a constant, a product of x,y,z. No additional conditions given.
$endgroup$
– LeoBonhart
Jan 2 at 14:17
$begingroup$
Oh yea? Find the minimum for k = 0. Also take a look at k = -1.
$endgroup$
– William Elliot
Jan 2 at 20:34
1
$begingroup$
There is no solution for k = 0. @LeoBonhart
$endgroup$
– William Elliot
Jan 4 at 20:37
|
show 1 more comment
0
0
0
$begingroup$
Exercise asks to verify that the sum of three quantities x, y, z, whose product is a constant k, is minimum when these three quantities are equal.
This is my amateurish attempt:
$x + y + z = S$;
$x*y*z=k$;
$z = frac{k}{xy}$;- x + y + $frac{k}{xy}$ = S;
$frac{partial S}{partial x} = 1 - frac{k}{x^2y} = frac{x^2y-k}{x^2y}$ and symmetricly $frac{partial S}{partial y} = frac{y^2x-k}{y^2x}$;
$x^2y-k = 0 rightarrow x^2y=k rightarrow x^2y=xyz rightarrow frac{x^2y}{xy}=z rightarrow x=z$ and symmetricly $y=z$, so $y=x=z$;- This part is doubtful to me. Proving that S is minimal when $x=y=z rightarrow$ $frac{partial ^2 S}{partial x^2} = frac{2k}{yx^3} rightarrow frac{2(xyz)}{yx^3} rightarrow frac{2z}{x^2} rightarrow frac{2x}{x^2} rightarrow frac{2}{x}$ and symmetricly $frac{2}{y}$;
Conclusion - if second derivative is positive, that means $S$ must be a minimum.
Please be free to comment if this solution is flawed or missing something.
calculus optimization
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
$endgroup$
Exercise asks to verify that the sum of three quantities x, y, z, whose product is a constant k, is minimum when these three quantities are equal.
This is my amateurish attempt:
$x + y + z = S$;
$x*y*z=k$;
$z = frac{k}{xy}$;- x + y + $frac{k}{xy}$ = S;
$frac{partial S}{partial x} = 1 - frac{k}{x^2y} = frac{x^2y-k}{x^2y}$ and symmetricly $frac{partial S}{partial y} = frac{y^2x-k}{y^2x}$;
$x^2y-k = 0 rightarrow x^2y=k rightarrow x^2y=xyz rightarrow frac{x^2y}{xy}=z rightarrow x=z$ and symmetricly $y=z$, so $y=x=z$;- This part is doubtful to me. Proving that S is minimal when $x=y=z rightarrow$ $frac{partial ^2 S}{partial x^2} = frac{2k}{yx^3} rightarrow frac{2(xyz)}{yx^3} rightarrow frac{2z}{x^2} rightarrow frac{2x}{x^2} rightarrow frac{2}{x}$ and symmetricly $frac{2}{y}$;
Conclusion - if second derivative is positive, that means $S$ must be a minimum.
Please be free to comment if this solution is flawed or missing something.
calculus optimization
calculus optimization
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
asked Jan 2 at 12:55
LeoBonhartLeoBonhart
134
134
$begingroup$
Did the book mention the Inequality of arithmetic and geometric means?
$endgroup$
– A.Γ.
Jan 2 at 13:18
$begingroup$
Were any conditions placed on k?
$endgroup$
– William Elliot
Jan 2 at 13:44
$begingroup$
@WilliamElliot k is just a constant, a product of x,y,z. No additional conditions given.
$endgroup$
– LeoBonhart
Jan 2 at 14:17
$begingroup$
Oh yea? Find the minimum for k = 0. Also take a look at k = -1.
$endgroup$
– William Elliot
Jan 2 at 20:34
1
$begingroup$
There is no solution for k = 0. @LeoBonhart
$endgroup$
– William Elliot
Jan 4 at 20:37
|
show 1 more comment
$begingroup$
Did the book mention the Inequality of arithmetic and geometric means?
$endgroup$
– A.Γ.
Jan 2 at 13:18
$begingroup$
Were any conditions placed on k?
$endgroup$
– William Elliot
Jan 2 at 13:44
$begingroup$
@WilliamElliot k is just a constant, a product of x,y,z. No additional conditions given.
$endgroup$
– LeoBonhart
Jan 2 at 14:17
$begingroup$
Oh yea? Find the minimum for k = 0. Also take a look at k = -1.
$endgroup$
– William Elliot
Jan 2 at 20:34
1
$begingroup$
There is no solution for k = 0. @LeoBonhart
$endgroup$
– William Elliot
Jan 4 at 20:37
$begingroup$
Did the book mention the Inequality of arithmetic and geometric means?
$endgroup$
– A.Γ.
Jan 2 at 13:18
$begingroup$
Did the book mention the Inequality of arithmetic and geometric means?
$endgroup$
– A.Γ.
Jan 2 at 13:18
$begingroup$
Were any conditions placed on k?
$endgroup$
– William Elliot
Jan 2 at 13:44
$begingroup$
Were any conditions placed on k?
$endgroup$
– William Elliot
Jan 2 at 13:44
$begingroup$
@WilliamElliot k is just a constant, a product of x,y,z. No additional conditions given.
$endgroup$
– LeoBonhart
Jan 2 at 14:17
$begingroup$
@WilliamElliot k is just a constant, a product of x,y,z. No additional conditions given.
$endgroup$
– LeoBonhart
Jan 2 at 14:17
$begingroup$
Oh yea? Find the minimum for k = 0. Also take a look at k = -1.
$endgroup$
– William Elliot
Jan 2 at 20:34
$begingroup$
Oh yea? Find the minimum for k = 0. Also take a look at k = -1.
$endgroup$
– William Elliot
Jan 2 at 20:34
1
1
$begingroup$
There is no solution for k = 0. @LeoBonhart
$endgroup$
– William Elliot
Jan 4 at 20:37
$begingroup$
There is no solution for k = 0. @LeoBonhart
$endgroup$
– William Elliot
Jan 4 at 20:37
|
show 1 more comment
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$begingroup$
Did the book mention the Inequality of arithmetic and geometric means?
$endgroup$
– A.Γ.
Jan 2 at 13:18
$begingroup$
Were any conditions placed on k?
$endgroup$
– William Elliot
Jan 2 at 13:44
$begingroup$
@WilliamElliot k is just a constant, a product of x,y,z. No additional conditions given.
$endgroup$
– LeoBonhart
Jan 2 at 14:17
$begingroup$
Oh yea? Find the minimum for k = 0. Also take a look at k = -1.
$endgroup$
– William Elliot
Jan 2 at 20:34
1
$begingroup$
There is no solution for k = 0. @LeoBonhart
$endgroup$
– William Elliot
Jan 4 at 20:37