Maximum value of a function with 2 variables
$begingroup$
Can someone help me finding maximum value of a ratio in quadratic function in 2 variables using proper mathematical methods.?
Question is as below.
If x and y are real numbers such that $x^2 -10x+y^2 +16=0$, determine the maximum value of the ratio $y/x$
I know there is Ramban method to solve this. Taking $y/x=k --> y=kx$ and forming equation in x , then applying $^2 - 4ac >=0$ for max min value of k.
Is there any way to using differentiation ?
Sorry in advance if this is a repeat. I am new to platform.
quadratics maxima-minima
edited Dec 12 '18 at 20:22
lomber
786420
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
$endgroup$
add a comment |
$begingroup$
Can someone help me finding maximum value of a ratio in quadratic function in 2 variables using proper mathematical methods.?
Question is as below.
If x and y are real numbers such that $x^2 -10x+y^2 +16=0$, determine the maximum value of the ratio $y/x$
I know there is Ramban method to solve this. Taking $y/x=k --> y=kx$ and forming equation in x , then applying $^2 - 4ac >=0$ for max min value of k.
Is there any way to using differentiation ?
Sorry in advance if this is a repeat. I am new to platform.
quadratics maxima-minima
edited Dec 12 '18 at 20:22
lomber
786420
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
$endgroup$
add a comment |
$begingroup$
Can someone help me finding maximum value of a ratio in quadratic function in 2 variables using proper mathematical methods.?
Question is as below.
If x and y are real numbers such that $x^2 -10x+y^2 +16=0$, determine the maximum value of the ratio $y/x$
I know there is Ramban method to solve this. Taking $y/x=k --> y=kx$ and forming equation in x , then applying $^2 - 4ac >=0$ for max min value of k.
Is there any way to using differentiation ?
Sorry in advance if this is a repeat. I am new to platform.
quadratics maxima-minima
edited Dec 12 '18 at 20:22
lomber
786420
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
$endgroup$
Can someone help me finding maximum value of a ratio in quadratic function in 2 variables using proper mathematical methods.?
Question is as below.
If x and y are real numbers such that $x^2 -10x+y^2 +16=0$, determine the maximum value of the ratio $y/x$
I know there is Ramban method to solve this. Taking $y/x=k --> y=kx$ and forming equation in x , then applying $^2 - 4ac >=0$ for max min value of k.
Is there any way to using differentiation ?
Sorry in advance if this is a repeat. I am new to platform.
quadratics maxima-minima
quadratics maxima-minima
edited Dec 12 '18 at 20:22
lomber
786420
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
edited Dec 12 '18 at 20:22
lomber
786420
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
edited Dec 12 '18 at 20:22
lomber
786420
edited Dec 12 '18 at 20:22
lomber
786420
edited Dec 12 '18 at 20:22
lomber
786420
786420
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
asked Dec 12 '18 at 19:32
Vishu SahniVishu Sahni
33
33
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that the equation is a circle with center $O(5,0)$ and radius $3$:
$$x^2 -10x+y^2 +16=0 iff (x-5)^2+y^2=9$$
The objective function is $frac yx=k iff y=kx$, whose contour lines will pass through the origin. So you need to find the slope of the tangent to the circle. See the graph:
$hspace{4cm}$
Hence, the slope is $k=frac 34$, which is the maximum value of $frac yx$ at $x=frac{16}{5}$ and $y=frac{12}{5}$.
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
$endgroup$
$begingroup$
Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks
$endgroup$
– Vishu Sahni
Dec 13 '18 at 21:40
add a comment |
$begingroup$
$y/x=m.$
$x^2-10x+(mx)^2+16=0.$
$(1+m^2)x^2 -10x+16=0.$
$small{(1+m^2)left (x^2-dfrac{10}{1+m^2}x right) +16=0.}$
Completing the square:
$small{(1+m^2)left (x-dfrac{5}{1+m^2}right)^2 -dfrac{25}{1+m^2}+16=0.}$
$small{(1+m^2)^2 left (x-dfrac{5}{1+m^2}right )^2 =-16(1+m^2)+25 ge 0.}$
Hence :
$25/16 ge 1+m^2$.
$9/16 ge m^2$.
Maximal $m:$
$m=3/4.$
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that the equation is a circle with center $O(5,0)$ and radius $3$:
$$x^2 -10x+y^2 +16=0 iff (x-5)^2+y^2=9$$
The objective function is $frac yx=k iff y=kx$, whose contour lines will pass through the origin. So you need to find the slope of the tangent to the circle. See the graph:
$hspace{4cm}$
Hence, the slope is $k=frac 34$, which is the maximum value of $frac yx$ at $x=frac{16}{5}$ and $y=frac{12}{5}$.
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
$endgroup$
$begingroup$
Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks
$endgroup$
– Vishu Sahni
Dec 13 '18 at 21:40
add a comment |
$begingroup$
Note that the equation is a circle with center $O(5,0)$ and radius $3$:
$$x^2 -10x+y^2 +16=0 iff (x-5)^2+y^2=9$$
The objective function is $frac yx=k iff y=kx$, whose contour lines will pass through the origin. So you need to find the slope of the tangent to the circle. See the graph:
$hspace{4cm}$
Hence, the slope is $k=frac 34$, which is the maximum value of $frac yx$ at $x=frac{16}{5}$ and $y=frac{12}{5}$.
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
$endgroup$
$begingroup$
Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks
$endgroup$
– Vishu Sahni
Dec 13 '18 at 21:40
add a comment |
$begingroup$
Note that the equation is a circle with center $O(5,0)$ and radius $3$:
$$x^2 -10x+y^2 +16=0 iff (x-5)^2+y^2=9$$
The objective function is $frac yx=k iff y=kx$, whose contour lines will pass through the origin. So you need to find the slope of the tangent to the circle. See the graph:
$hspace{4cm}$
Hence, the slope is $k=frac 34$, which is the maximum value of $frac yx$ at $x=frac{16}{5}$ and $y=frac{12}{5}$.
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
$endgroup$
Note that the equation is a circle with center $O(5,0)$ and radius $3$:
$$x^2 -10x+y^2 +16=0 iff (x-5)^2+y^2=9$$
The objective function is $frac yx=k iff y=kx$, whose contour lines will pass through the origin. So you need to find the slope of the tangent to the circle. See the graph:
$hspace{4cm}$
Hence, the slope is $k=frac 34$, which is the maximum value of $frac yx$ at $x=frac{16}{5}$ and $y=frac{12}{5}$.
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
answered Dec 12 '18 at 20:55
farruhotafarruhota
20.5k2739
20.5k2739
$begingroup$
Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks
$endgroup$
– Vishu Sahni
Dec 13 '18 at 21:40
add a comment |
$begingroup$
Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks
$endgroup$
– Vishu Sahni
Dec 13 '18 at 21:40
$begingroup$
Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks
$endgroup$
– Vishu Sahni
Dec 13 '18 at 21:40
$begingroup$
Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks
$endgroup$
– Vishu Sahni
Dec 13 '18 at 21:40
add a comment |
$begingroup$
$y/x=m.$
$x^2-10x+(mx)^2+16=0.$
$(1+m^2)x^2 -10x+16=0.$
$small{(1+m^2)left (x^2-dfrac{10}{1+m^2}x right) +16=0.}$
Completing the square:
$small{(1+m^2)left (x-dfrac{5}{1+m^2}right)^2 -dfrac{25}{1+m^2}+16=0.}$
$small{(1+m^2)^2 left (x-dfrac{5}{1+m^2}right )^2 =-16(1+m^2)+25 ge 0.}$
Hence :
$25/16 ge 1+m^2$.
$9/16 ge m^2$.
Maximal $m:$
$m=3/4.$
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$y/x=m.$
$x^2-10x+(mx)^2+16=0.$
$(1+m^2)x^2 -10x+16=0.$
$small{(1+m^2)left (x^2-dfrac{10}{1+m^2}x right) +16=0.}$
Completing the square:
$small{(1+m^2)left (x-dfrac{5}{1+m^2}right)^2 -dfrac{25}{1+m^2}+16=0.}$
$small{(1+m^2)^2 left (x-dfrac{5}{1+m^2}right )^2 =-16(1+m^2)+25 ge 0.}$
Hence :
$25/16 ge 1+m^2$.
$9/16 ge m^2$.
Maximal $m:$
$m=3/4.$
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$y/x=m.$
$x^2-10x+(mx)^2+16=0.$
$(1+m^2)x^2 -10x+16=0.$
$small{(1+m^2)left (x^2-dfrac{10}{1+m^2}x right) +16=0.}$
Completing the square:
$small{(1+m^2)left (x-dfrac{5}{1+m^2}right)^2 -dfrac{25}{1+m^2}+16=0.}$
$small{(1+m^2)^2 left (x-dfrac{5}{1+m^2}right )^2 =-16(1+m^2)+25 ge 0.}$
Hence :
$25/16 ge 1+m^2$.
$9/16 ge m^2$.
Maximal $m:$
$m=3/4.$
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$y/x=m.$
$x^2-10x+(mx)^2+16=0.$
$(1+m^2)x^2 -10x+16=0.$
$small{(1+m^2)left (x^2-dfrac{10}{1+m^2}x right) +16=0.}$
Completing the square:
$small{(1+m^2)left (x-dfrac{5}{1+m^2}right)^2 -dfrac{25}{1+m^2}+16=0.}$
$small{(1+m^2)^2 left (x-dfrac{5}{1+m^2}right )^2 =-16(1+m^2)+25 ge 0.}$
Hence :
$25/16 ge 1+m^2$.
$9/16 ge m^2$.
Maximal $m:$
$m=3/4.$
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
edited Dec 12 '18 at 23:43
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
answered Dec 12 '18 at 23:19
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
add a comment |
add a comment |
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