Find the range of $xy$ under the conditions: $x^2-xy+y^2=9$ and $|x^2-y^2|<9$
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Assume $x,y in mathbb{R}^+$, and satisfied the following expression:
$$x^2-xy+y^2=9$$
$$left|x^2-y^2right| < 9$$
find the range of $xy$
My approach: $x^2-xy+y^2=9$ $Rightarrow$ $xy+9=x^2+y^2 geq 2xy$ $Rightarrow$ $xy leq 9$
But I don't know how to find the lower bound. please help me..thanks very much.
algebra-precalculus analysis
asked Nov 26 at 2:49
Ginkgo
82
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up vote
1
down vote
favorite
Assume $x,y in mathbb{R}^+$, and satisfied the following expression:
$$x^2-xy+y^2=9$$
$$left|x^2-y^2right| < 9$$
find the range of $xy$
My approach: $x^2-xy+y^2=9$ $Rightarrow$ $xy+9=x^2+y^2 geq 2xy$ $Rightarrow$ $xy leq 9$
But I don't know how to find the lower bound. please help me..thanks very much.
algebra-precalculus analysis
asked Nov 26 at 2:49
Ginkgo
82
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Assume $x,y in mathbb{R}^+$, and satisfied the following expression:
$$x^2-xy+y^2=9$$
$$left|x^2-y^2right| < 9$$
find the range of $xy$
My approach: $x^2-xy+y^2=9$ $Rightarrow$ $xy+9=x^2+y^2 geq 2xy$ $Rightarrow$ $xy leq 9$
But I don't know how to find the lower bound. please help me..thanks very much.
algebra-precalculus analysis
asked Nov 26 at 2:49
Ginkgo
82
Assume $x,y in mathbb{R}^+$, and satisfied the following expression:
$$x^2-xy+y^2=9$$
$$left|x^2-y^2right| < 9$$
find the range of $xy$
My approach: $x^2-xy+y^2=9$ $Rightarrow$ $xy+9=x^2+y^2 geq 2xy$ $Rightarrow$ $xy leq 9$
But I don't know how to find the lower bound. please help me..thanks very much.
algebra-precalculus analysis
algebra-precalculus analysis
asked Nov 26 at 2:49
Ginkgo
82
asked Nov 26 at 2:49
Ginkgo
82
asked Nov 26 at 2:49
Ginkgo
82
asked Nov 26 at 2:49
Ginkgo
82
asked Nov 26 at 2:49
Ginkgo
82
82
add a comment |
add a comment |
1 Answer
1
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up vote
3
down vote
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For the lower bound you can use the follwing facts:
- $x^2-xy+y^2 = (x-y)^2 +xy Rightarrow 9-xy = (x-y)^2$
$x^2-xy+y^2 = (x+y)^2 -3xy Rightarrow 9+3xy = (x+y)^2$- $left|x^2-y^2right| < 9 Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 Leftrightarrow (9-xy)(9+3xy) < 81 Leftrightarrow xy(6-xy) < 0 stackrel{x,y >0}{Leftrightarrow} boxed{xy>6}$$
Together with your upper bound $boxed{xy leq 9}$ you get
$$boxed{6 < xy leq 9}$$
answered Nov 26 at 4:50
trancelocation
8,6421520
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For the lower bound you can use the follwing facts:
- $x^2-xy+y^2 = (x-y)^2 +xy Rightarrow 9-xy = (x-y)^2$
$x^2-xy+y^2 = (x+y)^2 -3xy Rightarrow 9+3xy = (x+y)^2$- $left|x^2-y^2right| < 9 Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 Leftrightarrow (9-xy)(9+3xy) < 81 Leftrightarrow xy(6-xy) < 0 stackrel{x,y >0}{Leftrightarrow} boxed{xy>6}$$
Together with your upper bound $boxed{xy leq 9}$ you get
$$boxed{6 < xy leq 9}$$
answered Nov 26 at 4:50
trancelocation
8,6421520
add a comment |
up vote
3
down vote
accepted
For the lower bound you can use the follwing facts:
- $x^2-xy+y^2 = (x-y)^2 +xy Rightarrow 9-xy = (x-y)^2$
$x^2-xy+y^2 = (x+y)^2 -3xy Rightarrow 9+3xy = (x+y)^2$- $left|x^2-y^2right| < 9 Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 Leftrightarrow (9-xy)(9+3xy) < 81 Leftrightarrow xy(6-xy) < 0 stackrel{x,y >0}{Leftrightarrow} boxed{xy>6}$$
Together with your upper bound $boxed{xy leq 9}$ you get
$$boxed{6 < xy leq 9}$$
answered Nov 26 at 4:50
trancelocation
8,6421520
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For the lower bound you can use the follwing facts:
- $x^2-xy+y^2 = (x-y)^2 +xy Rightarrow 9-xy = (x-y)^2$
$x^2-xy+y^2 = (x+y)^2 -3xy Rightarrow 9+3xy = (x+y)^2$- $left|x^2-y^2right| < 9 Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 Leftrightarrow (9-xy)(9+3xy) < 81 Leftrightarrow xy(6-xy) < 0 stackrel{x,y >0}{Leftrightarrow} boxed{xy>6}$$
Together with your upper bound $boxed{xy leq 9}$ you get
$$boxed{6 < xy leq 9}$$
answered Nov 26 at 4:50
trancelocation
8,6421520
For the lower bound you can use the follwing facts:
- $x^2-xy+y^2 = (x-y)^2 +xy Rightarrow 9-xy = (x-y)^2$
$x^2-xy+y^2 = (x+y)^2 -3xy Rightarrow 9+3xy = (x+y)^2$- $left|x^2-y^2right| < 9 Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 Leftrightarrow (9-xy)(9+3xy) < 81 Leftrightarrow xy(6-xy) < 0 stackrel{x,y >0}{Leftrightarrow} boxed{xy>6}$$
Together with your upper bound $boxed{xy leq 9}$ you get
$$boxed{6 < xy leq 9}$$
answered Nov 26 at 4:50
trancelocation
8,6421520
edited Nov 26 at 5:28
answered Nov 26 at 4:50
trancelocation
8,6421520
answered Nov 26 at 4:50
trancelocation
8,6421520
answered Nov 26 at 4:50
trancelocation
8,6421520
8,6421520
add a comment |
add a comment |
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