Maximum and minimum absolute value of a complex number

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Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$

I only have a vague idea to attack this problem.
Here's my thinking :

Let $z=a+bi$

Exploiting the fact that, $a^2+b^2=4$

We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$

So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$

The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$

Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$

Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.

I don't know if it's okay to set $b=0$ since $z$ would become a real number then.

edited 4 hours ago

greedoid

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asked 7 hours ago

ARahman

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up vote
3
down vote

favorite

Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$

I only have a vague idea to attack this problem.
Here's my thinking :

Let $z=a+bi$

Exploiting the fact that, $a^2+b^2=4$

We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$

So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$

The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$

Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$

Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.

I don't know if it's okay to set $b=0$ since $z$ would become a real number then.

edited 4 hours ago

greedoid

31k94187

asked 7 hours ago

ARahman

1085

add a commentÂ |Â

up vote
3
down vote

favorite

Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$

I only have a vague idea to attack this problem.
Here's my thinking :

Let $z=a+bi$

Exploiting the fact that, $a^2+b^2=4$

We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$

So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$

The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$

Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$

Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.

I don't know if it's okay to set $b=0$ since $z$ would become a real number then.

edited 4 hours ago

greedoid

31k94187

asked 7 hours ago

ARahman

1085

Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$

I only have a vague idea to attack this problem.
Here's my thinking :

Let $z=a+bi$

Exploiting the fact that, $a^2+b^2=4$

We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$

So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$

The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$

Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$

Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.

I don't know if it's okay to set $b=0$ since $z$ would become a real number then.

algebra-precalculus complex-numbers maxima-minima

edited 4 hours ago

greedoid

31k94187

asked 7 hours ago

ARahman

1085

edited 4 hours ago

greedoid

31k94187

asked 7 hours ago

ARahman

1085

edited 4 hours ago

greedoid

31k94187

edited 4 hours ago

greedoid

31k94187

edited 4 hours ago

greedoid

31k94187

asked 7 hours ago

ARahman

1085

asked 7 hours ago

ARahman

1085

asked 7 hours ago

ARahman

1085

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6 Answers
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A (very) faster way:

We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+frac12=frac52$$
Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.

Can you deal with the minimum now?

answered 6 hours ago

ajotatxe

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2
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$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$

answered 6 hours ago

Henry Lee

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2
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You could try like this (with help of triangle inequality):

$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$

clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.

edited 6 hours ago

answered 6 hours ago

greedoid

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up vote
1
down vote

Your idea is good, but you lose yourself in some computations (the maximum is correct, though).

Consider the square of the modulus:
$$
f(z)=left|z-frac1zright|^2=fracz^2-1^2
$$
Since $|z|=2$ by assumption, we can as well consider
$$
g(z)=|z^2-1|^2=(z^2-1)(barz^2-1)=z^2barz^2-z^2-barz^2+1=5-z^2-barz^2
$$
If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $barz^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then
$$
g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2
$$
The maximum is for $a=0$, the minimum for $a=pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.

We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.

answered 6 hours ago

egreg

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$ z - frac1z = fracz^2-1z $

So that, we get $left| z - frac1z right| $ = $fracz^2-12 $

If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:

$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $

Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $

We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $

$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $

So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.

Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$

if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$

answered 6 hours ago

Dominik Kutek

1013

New contributor

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Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]

Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.

So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.

Intuitively[2] I'd say that as $frac 2516 > frac 916$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.

As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.

So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.

====

[1]. There probably no need for this, but we can create an identity:

$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.

[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.

Actually formally was easier than my intuition.

edited 4 hours ago

answered 4 hours ago

fleablood

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

up vote
3
down vote

A (very) faster way:

Can you deal with the minimum now?

answered 6 hours ago

ajotatxe

50.5k13185

add a commentÂ |Â

up vote
3
down vote

A (very) faster way:

Can you deal with the minimum now?

answered 6 hours ago

ajotatxe

50.5k13185

add a commentÂ |Â

up vote
3
down vote

A (very) faster way:

Can you deal with the minimum now?

answered 6 hours ago

ajotatxe

50.5k13185

A (very) faster way:

Can you deal with the minimum now?

answered 6 hours ago

ajotatxe

50.5k13185

answered 6 hours ago

ajotatxe

50.5k13185

answered 6 hours ago

ajotatxe

50.5k13185

answered 6 hours ago

ajotatxe

50.5k13185

add a commentÂ |Â

up vote
2
down vote

$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$

answered 6 hours ago

Henry Lee

1,23816

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up vote
2
down vote

$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$

answered 6 hours ago

Henry Lee

1,23816

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up vote
2
down vote

$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$

answered 6 hours ago

Henry Lee

1,23816

$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$

answered 6 hours ago

Henry Lee

1,23816

answered 6 hours ago

Henry Lee

1,23816

answered 6 hours ago

Henry Lee

1,23816

answered 6 hours ago

Henry Lee

1,23816

add a commentÂ |Â

up vote
2
down vote

You could try like this (with help of triangle inequality):

$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$

clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.

edited 6 hours ago

answered 6 hours ago

greedoid

31k94187

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up vote
2
down vote

You could try like this (with help of triangle inequality):

$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$

clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.

edited 6 hours ago

answered 6 hours ago

greedoid

31k94187

add a commentÂ |Â

up vote
2
down vote

You could try like this (with help of triangle inequality):

$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$

clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.

edited 6 hours ago

answered 6 hours ago

greedoid

31k94187

You could try like this (with help of triangle inequality):

$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$

clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.

edited 6 hours ago

answered 6 hours ago

greedoid

31k94187

edited 6 hours ago

answered 6 hours ago

greedoid

31k94187

answered 6 hours ago

greedoid

31k94187

answered 6 hours ago

greedoid

31k94187

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up vote
1
down vote

Your idea is good, but you lose yourself in some computations (the maximum is correct, though).

We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.

answered 6 hours ago

egreg

169k1283191

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up vote
1
down vote

Your idea is good, but you lose yourself in some computations (the maximum is correct, though).

We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.

answered 6 hours ago

egreg

169k1283191

add a commentÂ |Â

up vote
1
down vote

Your idea is good, but you lose yourself in some computations (the maximum is correct, though).

We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.

answered 6 hours ago

egreg

169k1283191

Your idea is good, but you lose yourself in some computations (the maximum is correct, though).

We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.

answered 6 hours ago

egreg

169k1283191

answered 6 hours ago

egreg

169k1283191

answered 6 hours ago

egreg

169k1283191

answered 6 hours ago

egreg

169k1283191

add a commentÂ |Â

up vote
0
down vote

$ z - frac1z = fracz^2-1z $

So that, we get $left| z - frac1z right| $ = $fracz^2-12 $

If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:

$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $

Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $

We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $

$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $

So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.

Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$

if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$

answered 6 hours ago

Dominik Kutek

1013

New contributor

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up vote
0
down vote

$ z - frac1z = fracz^2-1z $

So that, we get $left| z - frac1z right| $ = $fracz^2-12 $

If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:

$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $

Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $

We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $

$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $

So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.

Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$

if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$

answered 6 hours ago

Dominik Kutek

1013

New contributor

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up vote
0
down vote

$ z - frac1z = fracz^2-1z $

So that, we get $left| z - frac1z right| $ = $fracz^2-12 $

If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:

$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $

Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $

We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $

$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $

So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.

Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$

if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$

answered 6 hours ago

Dominik Kutek

1013

New contributor

$ z - frac1z = fracz^2-1z $

So that, we get $left| z - frac1z right| $ = $fracz^2-12 $

If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:

$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $

Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $

We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $

$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $

So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.

Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$

if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$

answered 6 hours ago

Dominik Kutek

1013

New contributor

answered 6 hours ago

Dominik Kutek

1013

New contributor

answered 6 hours ago

Dominik Kutek

1013

answered 6 hours ago

Dominik Kutek

1013

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Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]

Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.

So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.

As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.

So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.

====

[1]. There probably no need for this, but we can create an identity:

$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.

[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.

Actually formally was easier than my intuition.

edited 4 hours ago

answered 4 hours ago

fleablood

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up vote
0
down vote

Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]

Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.

So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.

As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.

So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.

====

[1]. There probably no need for this, but we can create an identity:

$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.

[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.

Actually formally was easier than my intuition.

edited 4 hours ago

answered 4 hours ago

fleablood

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Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]

Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.

So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.

As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.

So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.

====

[1]. There probably no need for this, but we can create an identity:

$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.

[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.

Actually formally was easier than my intuition.

edited 4 hours ago

answered 4 hours ago

fleablood

Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]

Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.

So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.

As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.

So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.

====

[1]. There probably no need for this, but we can create an identity:

$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.

[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.

Actually formally was easier than my intuition.

edited 4 hours ago

answered 4 hours ago

fleablood

edited 4 hours ago

answered 4 hours ago

fleablood

answered 4 hours ago

fleablood

answered 4 hours ago

fleablood

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