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l is equal to the minimum value of the expression 2(y-2)^2 + 4(x -7)^2 + (y+4)^2 find [2018/l] where [] denotes the greatest integer function
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$l$ is equal to the minimum value of the expression $2(y-2)^2 + 4(x -7)^2 + (y+4)^2$. Find $[2018/l]$ where $[cdot]$ denotes the greatest integer function.
Here is my approach:
In order to obtain the minimum value of the expression, one of the terms has to be zero, so let $x=7$ and $y=2$.
We get $(2+4)^2=36$
and $[2018/36]=56$.
Is this answer right?
algebra-precalculus polynomials roots
edited Aug 10 at 6:09
user529760
686417
asked Aug 6 at 16:12
John Tom
816
add a comment |Â
up vote
-1
down vote
favorite
$l$ is equal to the minimum value of the expression $2(y-2)^2 + 4(x -7)^2 + (y+4)^2$. Find $[2018/l]$ where $[cdot]$ denotes the greatest integer function.
Here is my approach:
In order to obtain the minimum value of the expression, one of the terms has to be zero, so let $x=7$ and $y=2$.
We get $(2+4)^2=36$
and $[2018/36]=56$.
Is this answer right?
algebra-precalculus polynomials roots
edited Aug 10 at 6:09
user529760
686417
asked Aug 6 at 16:12
John Tom
816
Here is a counterexample to your claim: Find the minimum of $(y-2)^2+(y+2)^2$. According to your claim you suggest setting $y$ so that one of these terms is zero, e.g. the first, which would happen at $y=2$, giving a result of $(2-2)^2+(2+2)^2=0^2+4^2=16$, but for my example $y=0$ happens to give a smaller result: $(2-0)^2+(2+0)^2=2^2+2^2=8<16$. In general, finding minimums and maximums can be accomplished by using differential calculus (or results from calculus that are taught early).
– JMoravitz
Aug 6 at 16:30
Here's a MathJax tutorial :)
– Shaun
Aug 6 at 16:33
For the special case of parabolas, if written in the form $f(x)=a(x-h)^2+k$, the maximum/minimum (max if $a>0$ and min if $a<0$) occurs at the point $f(h)=k$. For parabolas written in general form $f(x)=ax^2+bx+c$ it occurs at $f(frac-b2a)$. This is sometimes taught well before derivatives.
– JMoravitz
Aug 6 at 16:35
Hint: $;2(y-2)^2 + (y+4)^2 = 3 y^2 + 24,$.
– dxiv
Aug 7 at 1:12
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$l$ is equal to the minimum value of the expression $2(y-2)^2 + 4(x -7)^2 + (y+4)^2$. Find $[2018/l]$ where $[cdot]$ denotes the greatest integer function.
Here is my approach:
In order to obtain the minimum value of the expression, one of the terms has to be zero, so let $x=7$ and $y=2$.
We get $(2+4)^2=36$
and $[2018/36]=56$.
Is this answer right?
algebra-precalculus polynomials roots
edited Aug 10 at 6:09
user529760
686417
asked Aug 6 at 16:12
John Tom
816
$l$ is equal to the minimum value of the expression $2(y-2)^2 + 4(x -7)^2 + (y+4)^2$. Find $[2018/l]$ where $[cdot]$ denotes the greatest integer function.
Here is my approach:
In order to obtain the minimum value of the expression, one of the terms has to be zero, so let $x=7$ and $y=2$.
We get $(2+4)^2=36$
and $[2018/36]=56$.
Is this answer right?
algebra-precalculus polynomials roots
edited Aug 10 at 6:09
user529760
686417
asked Aug 6 at 16:12
John Tom
816
edited Aug 10 at 6:09
user529760
686417
edited Aug 10 at 6:09
user529760
686417
edited Aug 10 at 6:09
user529760
686417
686417
asked Aug 6 at 16:12
John Tom
816
asked Aug 6 at 16:12
John Tom
816
asked Aug 6 at 16:12
John Tom
816
816
Here is a counterexample to your claim: Find the minimum of $(y-2)^2+(y+2)^2$. According to your claim you suggest setting $y$ so that one of these terms is zero, e.g. the first, which would happen at $y=2$, giving a result of $(2-2)^2+(2+2)^2=0^2+4^2=16$, but for my example $y=0$ happens to give a smaller result: $(2-0)^2+(2+0)^2=2^2+2^2=8<16$. In general, finding minimums and maximums can be accomplished by using differential calculus (or results from calculus that are taught early).
– JMoravitz
Aug 6 at 16:30
Here's a MathJax tutorial :)
– Shaun
Aug 6 at 16:33
For the special case of parabolas, if written in the form $f(x)=a(x-h)^2+k$, the maximum/minimum (max if $a>0$ and min if $a<0$) occurs at the point $f(h)=k$. For parabolas written in general form $f(x)=ax^2+bx+c$ it occurs at $f(frac-b2a)$. This is sometimes taught well before derivatives.
– JMoravitz
Aug 6 at 16:35
Hint: $;2(y-2)^2 + (y+4)^2 = 3 y^2 + 24,$.
– dxiv
Aug 7 at 1:12
add a comment |Â
Here is a counterexample to your claim: Find the minimum of $(y-2)^2+(y+2)^2$. According to your claim you suggest setting $y$ so that one of these terms is zero, e.g. the first, which would happen at $y=2$, giving a result of $(2-2)^2+(2+2)^2=0^2+4^2=16$, but for my example $y=0$ happens to give a smaller result: $(2-0)^2+(2+0)^2=2^2+2^2=8<16$. In general, finding minimums and maximums can be accomplished by using differential calculus (or results from calculus that are taught early).
– JMoravitz
Aug 6 at 16:30
Here's a MathJax tutorial :)
– Shaun
Aug 6 at 16:33
For the special case of parabolas, if written in the form $f(x)=a(x-h)^2+k$, the maximum/minimum (max if $a>0$ and min if $a<0$) occurs at the point $f(h)=k$. For parabolas written in general form $f(x)=ax^2+bx+c$ it occurs at $f(frac-b2a)$. This is sometimes taught well before derivatives.
– JMoravitz
Aug 6 at 16:35
Hint: $;2(y-2)^2 + (y+4)^2 = 3 y^2 + 24,$.
– dxiv
Aug 7 at 1:12
Here is a counterexample to your claim: Find the minimum of $(y-2)^2+(y+2)^2$. According to your claim you suggest setting $y$ so that one of these terms is zero, e.g. the first, which would happen at $y=2$, giving a result of $(2-2)^2+(2+2)^2=0^2+4^2=16$, but for my example $y=0$ happens to give a smaller result: $(2-0)^2+(2+0)^2=2^2+2^2=8<16$. In general, finding minimums and maximums can be accomplished by using differential calculus (or results from calculus that are taught early).
– JMoravitz
Aug 6 at 16:30
Here is a counterexample to your claim: Find the minimum of $(y-2)^2+(y+2)^2$. According to your claim you suggest setting $y$ so that one of these terms is zero, e.g. the first, which would happen at $y=2$, giving a result of $(2-2)^2+(2+2)^2=0^2+4^2=16$, but for my example $y=0$ happens to give a smaller result: $(2-0)^2+(2+0)^2=2^2+2^2=8<16$. In general, finding minimums and maximums can be accomplished by using differential calculus (or results from calculus that are taught early).
– JMoravitz
Aug 6 at 16:30
Here's a MathJax tutorial :)
– Shaun
Aug 6 at 16:33
Here's a MathJax tutorial :)
– Shaun
Aug 6 at 16:33
For the special case of parabolas, if written in the form $f(x)=a(x-h)^2+k$, the maximum/minimum (max if $a>0$ and min if $a<0$) occurs at the point $f(h)=k$. For parabolas written in general form $f(x)=ax^2+bx+c$ it occurs at $f(frac-b2a)$. This is sometimes taught well before derivatives.
– JMoravitz
Aug 6 at 16:35
For the special case of parabolas, if written in the form $f(x)=a(x-h)^2+k$, the maximum/minimum (max if $a>0$ and min if $a<0$) occurs at the point $f(h)=k$. For parabolas written in general form $f(x)=ax^2+bx+c$ it occurs at $f(frac-b2a)$. This is sometimes taught well before derivatives.
– JMoravitz
Aug 6 at 16:35
Hint: $;2(y-2)^2 + (y+4)^2 = 3 y^2 + 24,$.
– dxiv
Aug 7 at 1:12
Hint: $;2(y-2)^2 + (y+4)^2 = 3 y^2 + 24,$.
– dxiv
Aug 7 at 1:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Guide:
Nope, that is not the right way.
Letting $x=7$ is a good move.
now, let's focus on the $y$ terms. To minimize
$$2(y-2)^2+(y+4)^2$$
You can differentiate the function with respect to $y$ and equate it to zero to solve for the $y$ corresponding to the minimum value.
Note that the function is convex hence there is a minimum.
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Guide:
Nope, that is not the right way.
Letting $x=7$ is a good move.
now, let's focus on the $y$ terms. To minimize
$$2(y-2)^2+(y+4)^2$$
You can differentiate the function with respect to $y$ and equate it to zero to solve for the $y$ corresponding to the minimum value.
Note that the function is convex hence there is a minimum.
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
add a comment |Â
up vote
1
down vote
Guide:
Nope, that is not the right way.
Letting $x=7$ is a good move.
now, let's focus on the $y$ terms. To minimize
$$2(y-2)^2+(y+4)^2$$
You can differentiate the function with respect to $y$ and equate it to zero to solve for the $y$ corresponding to the minimum value.
Note that the function is convex hence there is a minimum.
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Guide:
Nope, that is not the right way.
Letting $x=7$ is a good move.
now, let's focus on the $y$ terms. To minimize
$$2(y-2)^2+(y+4)^2$$
You can differentiate the function with respect to $y$ and equate it to zero to solve for the $y$ corresponding to the minimum value.
Note that the function is convex hence there is a minimum.
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
Guide:
Nope, that is not the right way.
Letting $x=7$ is a good move.
now, let's focus on the $y$ terms. To minimize
$$2(y-2)^2+(y+4)^2$$
You can differentiate the function with respect to $y$ and equate it to zero to solve for the $y$ corresponding to the minimum value.
Note that the function is convex hence there is a minimum.
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
answered Aug 6 at 16:23
Siong Thye Goh
78k134897
78k134897
add a comment |Â
add a comment |Â
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Here is a counterexample to your claim: Find the minimum of $(y-2)^2+(y+2)^2$. According to your claim you suggest setting $y$ so that one of these terms is zero, e.g. the first, which would happen at $y=2$, giving a result of $(2-2)^2+(2+2)^2=0^2+4^2=16$, but for my example $y=0$ happens to give a smaller result: $(2-0)^2+(2+0)^2=2^2+2^2=8<16$. In general, finding minimums and maximums can be accomplished by using differential calculus (or results from calculus that are taught early).
– JMoravitz
Aug 6 at 16:30
Here's a MathJax tutorial :)
– Shaun
Aug 6 at 16:33
For the special case of parabolas, if written in the form $f(x)=a(x-h)^2+k$, the maximum/minimum (max if $a>0$ and min if $a<0$) occurs at the point $f(h)=k$. For parabolas written in general form $f(x)=ax^2+bx+c$ it occurs at $f(frac-b2a)$. This is sometimes taught well before derivatives.
– JMoravitz
Aug 6 at 16:35
Hint: $;2(y-2)^2 + (y+4)^2 = 3 y^2 + 24,$.
– dxiv
Aug 7 at 1:12