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Let $x= 4sinA, y= 4cosA$ . $z=5sinB, t= 5cosB$ .

$xt-yz=20(sin(B-A))$ .

$B-A=90$ .

$P= xz= 20cosAcosB = 10sin2B$ .

$Max: 10$ .

Solution 1 (Geometric)

$\frac{1}{2}(xt-yz)$ is the area of the triangle with vertices $(0,0),(x,y),(z,t)$ . But since $(x,y)$ lies on a circle with radius 4 and $(z,t)$ lies on a circle with radius 5, this area is less than or equal to $\frac{1}{2}\times 5 \times 4=10$ with equality if $(x,y)$ and $(z,t)$ are perpendicular. We have equality so we can write $x=4\cos \theta$ and $z=-5\sin \theta$ and $x z=-10\sin 2\theta$ which has a maximum of 10.

Solution 2 (Algebraic)

Cauchy-Schwarz gives $xt-yz\leq\sqrt{x^{2}+y^{2}}\sqrt{z^{2}+y^{2}}=20$ , since we have equality we have $t=\lambda x, z=-\lambda y$ and $x z=\frac{zt}{\lambda}$ . Substituting into $z^{2}+t^{2}=25$ gives $\lambda^{2}=\frac{25}{16}$ and AM-GM gives $zt\leq\frac{25}{2}$ . It is easy to show from this that $xz\leq 10$

In both cases we can obtain equality using the points $(x,y)=(\frac{4}{\sqrt{2}}, -\frac{4}{\sqrt{2}})$ $(z,t)=(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$

Let $\vec{u}=(x,y)$ and $\vec{v}=(t,-z)$ . Then the three given equations are $\|\vec{u}\|^2=16 \\ \|\vec{v}\|^2=25 \\ \vec{u} \cdot \vec{v} = 20$ Since $20=\vec{u} \cdot \vec{v} = \|\vec{u}\|\|\vec{v}\| \cos \theta=4 \cdot 5 \cos \theta$ where $\theta$ is the angle between the vectors, we have $\cos \theta = 1$ and the vectors are co-directional. That is, $\vec{v}= \frac{5}{4}\vec{u}$ .

We are to find the maximum of $P=xz=- \frac{5}{4} xy$ given that $x$ and $y$ are on the circle $x^2+y^2=16$ . This will occur when $x=-y=\pm 2\sqrt{2}$ and $P=10$ .

This is my process, which may be a bit long due to unnecessary steps.

$16 \times 25 = 20^2$ so $(x^2+y^2)(z^2+t^2) = (xt-yz)^2$

Expanding gives $(xz)^2 + (xt)^2 + (yz)^2 + (yt)^2 = (xt)^2 + (yz)^2 - 2xyzt$ $(xz)^2 + (yt)^2 = -2xyzt$

LHS is positive so RHS must be positive as well, so one or three of the variables are negative. Looking at $xt-yz=20$ , one of $y, z$ is negative and either both of $x, t$ are positive or negative, and flipping their signs won't change the equations so WLOG let both of them be positive. Flip the sign of the other negative variable to get $xt+yz=20$ as the new third equation. The other two equations aren't affected. Now all variables are positive.

Using AM-GM on $\frac{(xz)^2 + (yt)^2}{2}$ gives $(xz)^2 + (yt)^2 \ge 2xyzt$ but it was given that $(xz)^2 + (yt)^2 = 2xyzt$ (negative sign was removed because of new equations) so $(xz)^2 = (yt)^2$ due to equality and $xz = yt$ because both are positive.

Let $xz = yt = a$ . The first two equations gives $\frac{x}{z} + \frac{y}{t} = \frac{16}{a}$ and $\frac{z}{x} + \frac{t}{y} = \frac{16}{a}$ . Dividing the equations gives $\frac{xy}{zt} = \frac{16}{25}$ (I found this by letting $a = \frac{x}{z}$ and $b = \frac{y}{t}$ ). Dividing $a$ from both sides of $25xy = 16zt$ gives $25\frac{x}{t} = 16\frac{t}{x}$ and $25\frac{y}{z} = 16\frac{z}{y}$ or $5x=4t$ and $5y = 4z$ .

Using AM-GM on the first equation gives $xy \le 8$ . Substituting $y = \frac{4}{5}z$ gives $xz \le \boxed{10}$ .

From Lagrange's identity; $(x^{2}+y^{2})(z^{2}+t^{2})=(xt-yz)^{2} + (xz+yt)^{2}$ .

Substituting we get $xz+yt = 0 \rightarrow xz = -yt$

From equation 1 and 2; using AM-GM we get

$400 = (x^{2}+y^{2})(z^{2}+t^{2}) \geq 4\sqrt{x^{2}y^{2}z^{2}t^{2}} = 4|xyzt| = 4|-(xz)^{2}| = 4(xz)^{2}$

Therefore, $xz \leq \boxed{10}$ ~~~

I solved this using complex numbers.

From the first two equations, we have $|x+yi| = 4$ and $|t+zi| = 5$ . Therefore, $4*5 = \left|(x+yi)(t+zi) \right| = |(xt-yz) + (xz+yt)i| = |20 + (xz+yt)i|$

Thus, $|20 + (xz+yt)i| = 20$ . Therefore, we must have $xz+yt = 0$ . Therefore, $(x+yi)(t+zi) = 20$ .

Since $|x+yi| = 4$ and $|t+zi| = 5$ , there must exist real numbers $a,b$ such that $x+yi = 4e^{ia}$ and $t+zi = 5e^{ib}$ . Their product is $20e^{i(a+b)}$ which must equal 20. Hence $b = -a$ . Using Euler's formula, we write $x = 4cos(a)$ and $z = 5sin(b) = -5sin(a)$ . Therefore $xz = -20cos(a)sin(a) = -10sin(2a)$ . This has a maximum value of $\boxed{10}$ .

Not by guessing,

x t = 20 + y z

Squaring both sides. With x^2 = 16 - y^2 and t^2 = 25 - z^2,

x^2 t^2 = 400 + 40 y z + y^2 z^2

=> (16 - y^2)(25 - z^2) = 400 + 40 y z + y^2 z^2

=> 400 - 25 y^2 - 16 z^2 + y^2 z^2 = 400 + 40 y z + y^2 z^2

=> 25 y^2 + 40 y z + 16 z^2 = 0

=> (5 y + 4 z)^2 = 0

=> z = (- 5/ 4) y

Since x = +/- sqrt(16 - y^2),

Maximum x z = (5/ 4) sqrt(16 y^2 - y^4);

Let f = 16 y^2 - y^4,

d f/ d y = 32 y - 4 y^3 and d^2 f/ d y^2 = 32 - 12 y^2

d f / d y = 0 => y^2 = 8 where d^2 f/ d y^2 = -64 < 0 {Indicates a maximum}

Maximum x z = (5/ 4) sqrt(16 x 8 - 64) = 10 where

[t, x, y, z] = [5/ sqrt(2), 2 sqrt(2), - 2 sqrt(2), 5/ sqrt(2)] OR [- 5/ sqrt(2), - 2 sqrt(2), 2 sqrt(2), - 5/ sqrt(2)] which satisfies 3 initial equations.

My solution got me the correct answer but I don't know why it works... make x^2=y^2, z^2=t^2, then we have (25/2)^(1/2) and 8^(1/2), when multiplied, gives 10.

I have a simplistic solution, which gave me the correct answer, but it might be incorrect, since I did not even need the 3rd equation. So how I solved this was:

X^{2} + Y^{2} =16 ---- (1) Perform partial differentiation to this, to get: 2X+2Y=0. Equate the left hand side to 0, to arrive at Y = -X. Substitute this in equation (1), and solve for X. We get X = 2 multiplied by Root of 2. This is the maximum value of X.

Perform similarly for z^{2} + t^{2} =25, solve for z, to arrive at the maximum value of z as 5 / Root of 2.

Multiply these two maximum values of x and z:

{2 . (Root of 2 )} multiplied by { 5 / Root of 2 } We get 10.

I got that $xz \leq 41/4$ . Here's my work... Notice how 16,25,and 20 have a relationship: $\sqrt{16} \times \sqrt{25}=20$ Therefore, we can combine these three equations into $\sqrt{x^{2}+y^{2}} \sqrt{z^{2}+t^{2}}=xt-yz$ Simplifying gives $x^{2}z^{2}+2xzyt+y^{2}t^{2}=0$ -> $xz+yt=0$ Now we add the first two equations together. Using our previous finding and factoring the difference of squares we get $(x+z-\sqrt{41})(x+z+\sqrt{41})=-(y-z)^{2}$ Since the right side is negative, the only way for this equation to be true is if $-\sqrt{41} \leq x+z \leq \sqrt{41}$ For xz to be maximized, we look at the cases $x+z=\sqrt{41}$ and $x+z=-\sqrt{41}$ . It is pretty easy to see that that they will both yield the same maximum value of xz since the xs and zs yielding the maximum value of xz in each case will just be negatives of each other. Therefore, we can just look at $x+z=\sqrt{41}$ and assume that $x$ and $z$ are non-negative real numbers to use AM-GM. $\sqrt{xz} \leq \frac{x+z}{2}$ -> $\sqrt{xz} \leq \frac{\sqrt{41}}{2}$ -> $xz \leq 41/4$

I think I found simple algebra in it.

eq. 1 $\times$ eq. 2 $\Rightarrow$ $x^{2}z^{2}+y^{2}t^{2}+x^{2}t^{2}+y^{2}t^{2}=400 \quad \cdots (4)$ squaring eq. 3 $\Rightarrow$ $x^{2}t^{2}+y^{2}z^{2}-2xyzt=400 \quad \cdots (5)$ $(4)-(5)\Rightarrow$ $x^{2}z^{2}+y^{2}t^{2}+2xyzt=0$ $\Rightarrow xz+yt=0$ $\Rightarrow \boxed{xz=-yt}$ Also from (1) and (2) we've got $y=\sqrt{16-x^2}$ and $t=\sqrt{25-t^2}$
$\therefore x^{2}z^{2}=y^{2}t^{2}=400-25x^{2}-16z^{2}+x^{2}z^{2}$ $\Rightarrow (5x)^{2}+(4z)^{2}=20^2$ $\therefore z=\frac{5\sqrt{16-x^2}}{4}$ $\therefore xz=\frac{5}{4}\times\sqrt{x^{2}(16-x^{2})=64-(x+8)^{2}}$ $\Rightarrow\boxed{\text{max}(xz)=10}$

Well....we could try to put all variables depended upon a particular parameter....and get an equation whose maximum value is required.....well to put it in simpler words....consider x=4cos(k) y=-4sin(k), Z=5sin(k) and and t=5cos(k)...these paramteric substitution satisfy all the given three equations...and we need to maximize P=20 sin(k)cos(k)....which should be a piece of cake....sorry for the dull solution though.... I hope it helps

take complex no. method let H = x+iy and G=z-it Use these modulus{H+G}< mod{H}+mod{G}...............................................1 HG={xz+yt}+i{yz-xt}.............................2 just use above eq.

From 1st and 2nd equations we get:-
$-4\le x,y \le 4................ -5\le t,z \le 5.\\To ~satisfy~equation~3......\\ (x,y,z,t) =(4,00,5) ~~or~~=(0,4,5,0)~~when ~~P= 0.\\$
To obtain extreme values, we can create symetray by....
$x^2 = y^2~~and~~t^2=z^2 ~~and~to ~satisfy~3rd~equation \\ we~let~~y=-x. \\Then~~(x,y,z,t) =(2*\sqrt2, 2*\sqrt2,~ 5/ \sqrt2, ~5/ \sqrt2)~~and~~ P=10. \\ ~~\\$
{ I am not satisfied with my arguments. Can any one improve ?}