Maximum Manipulation

For non geometric solution :

$\cfrac { 1 }{ { x }^{ 2 } } +\cfrac { 1 }{ { y }^{ 2 } } \quad =\quad 1\quad \quad (\quad \because \quad x\quad ,\quad y\quad \neq \quad 0\quad )$ .

$x\quad =\quad \sec { \theta } \quad \quad \& \quad \quad y\quad =\quad \csc { \theta }$ .

$E\quad =\cfrac { 5x\quad +\quad 12y\quad +\quad 7xy }{ xy } \quad \\ \\ E\quad =\quad \frac { 5 }{ y } +\frac { 12 }{ x } \quad +\quad 7\\ \\ E\quad =\quad 5\sin { \theta } \quad +\quad 12\cos { \theta } \quad +\quad 7\\ \\ { E }_{ max }\quad =\quad 13\quad +\quad 7\quad =\quad 20$ .

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But I believe There must be an geometric Solution ! So I'am working on it when I got correct then I will Surely Post it !

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Geometric solution: Triangle

Let denote these by standard notation of Triangle ABC

$x\quad =\quad a\quad ,\quad y\quad =\quad b\quad ,\quad xy\quad =\quad c$ .

So $E\quad =\cfrac { 5a\quad +\quad 12b\quad +\quad 7c }{ c } \quad \\ E\quad =\quad \cfrac { 5\sin { A } \quad +\quad 12\sin { B } \quad +\quad 7\sin { C } }{ \sin { C } } \\ \\ E\quad =\quad \cfrac { 5\sin { A } \quad +\quad 12\sin { B } \quad +\quad 7 }{ 1 } \quad \quad (\because \quad \sin { C } \quad =\quad 1\quad )\\ \\ { E }_{ max }=\quad 20$ .

if $x,y>0 .$ by cauchy : $13xy=\sqrt{(25+144)(x²+y²)} \geq 5x+12y$ so: $\frac{5x+12y+7xy}{xy}=\frac{5x+12y}{xy}+7\leq \frac{13xy}{xy}+7=20$

$Question\quad can\quad be\quad written\quad as\\ \left( \frac { 5 }{ y } +\frac { 12 }{ x } +7 \right) \\ We\quad would\quad ommit\quad 7\quad as\quad its\quad a\quad constant\\ \\ Applying\quad cauchy's\quad inequality\\ { \left( \frac { 5 }{ y } +\frac { 12 }{ x } \right) }^{ 2 }\le \left( 25+144 \right) (\frac { x^{ 2 }+{ y }^{ 2 } }{ x^{ 2 }y^{ 2 } } )\\ { \left( \frac { 5 }{ y } +\frac { 12 }{ x } \right) }\le 13\\ now\quad adding\quad 7\\ \because { \left( \frac { 5 }{ y } +\frac { 12 }{ x } +7 \right) }\le 20$

A calculus approach: (little longer)

From the equation $x^2+y^2=x^2y^2$ , we can get that $y= \frac {x}{ \sqrt {x^2-1}}$ . Solving with the other equation we get $\frac {5x+12 \frac {x}{ \sqrt {x^2-1}}+7 \frac {x^2}{ \sqrt {x^2-1}}}{ \frac {x^2}{ \sqrt {x^2-1}}}$

This (function) can be simplified to:

$\frac {5}{x} \sqrt {x^2-1}+ \frac {12}{x}+7$

Knowing that the derivative of the function will be equal to zero at its maximum value, after differentiation and some simplifications we get:

$12 \sqrt {x^2-1}=5$

Then: $x= \frac {12}{13}$ Then, the maximum value of the function will be at $x= \frac {12}{13}$ which is $\boxed {20}$

Substituting for sine and cosine would solve the problem. From $x^2 + y^2 = X^2 Y^2$ , we try to let $x^2 y^2$ be the square of radius. And $x$ and $y$ are now $R cos( \theta )$ and $R sin( \theta )$ respectively. Substituting, we obtain $(5 R cos( \theta ) + 12 R sin( \theta ) + 7 R ) / R$ , yielding $5 cos( \theta ) + 12 sin( \theta ) + 7$ . From Cauchy, we get the maximum of $5 cos( \theta ) + 12 sin( \theta )$ , which is square root of $[( 5^2+12^2)( cos^2 ( \theta ) + sin^2 ( \theta ) )]$ . Summing up, we get $13+7=20.$

Lagrange Multiplier's Method:

Let $f(x,y) = \dfrac{5x+12y+7xy}{xy} = \dfrac{5}{y} + \dfrac{12}{x} + 7$

The partial derivatives:

$f_x = -\dfrac{12}{x^2} \quad\quad f_y = -\dfrac{5}{y^2}$

Let $g(x,y) = x^2+y^2 - x^2y^2$

The partial derivatives:

$g_x = 2x - 2xy^2 \quad\quad g_y = 2y - 2x^2y$

Now, we can equate them:

$-\dfrac{12}{x^2} = \lambda (2x - 2xy^2) \implies$ Eq. (1)

$-\dfrac{5}{y^2} = \lambda (2y - 2x^2y) \implies$ Eq. (2)

Plus the original condition:

$x^2+y^2 = x^2y^2 \implies$ Eq. (3)

Now, Eq. (1) $\div$ Eq. (2):

$\dfrac{\left(-\dfrac{12}{x^2} \right)}{\left(-\dfrac{5}{y^2} \right)} = \dfrac{\lambda(2x - 2xy^2)}{\lambda (2y - 2x^2y)}\\ \dfrac{12y^2}{5x^2} = \dfrac{2x(1 - y^2)}{2y(1 - x^2)}\\ 12y^3(1-x^2) = 5x^3(1-y^2)$

From Eq. (3):

$x^2 + y^2 = x^2y^2 \implies y^2 = \dfrac{x^2}{x^2-1}$

Substitute it in:

$12 \left( \dfrac{x^2}{x^2-1} \right)^{3/2} (1-x^2) = 5x^3 \left(1-\dfrac{x^2}{x^2-1} \right)$

I don't like square roots, so I square the entire equation (even if you don't square it, you will still get the same answers):

$144 \left(\dfrac{x^2}{x^2-1} \right)^{3} (1-x^2)^2 = 25x^6 \left(\dfrac{-1}{x^2-1} \right)^2\\ \dfrac{144x^6(x^2-1)^2}{(x^2-1)^3} = \dfrac{25x^6}{(x^2-1)^2}\\ 144x^6(x^2-1)^4 = 25x^6(x^2-1)^3\\ 144x^6(x^2-1)^4 - 25x^6(x^2-1)^3 = 0\\ x^6(x^2-1)^3(144(x^2-1) - 25) = 0\\ x=0, x= \pm 1, x = \pm \dfrac{13}{12}$

Now, $x,y \neq 0$ , so we reject $x=0$

For $x = \pm 1 \implies x^2=1$ , we substitute it into Eq. (3):

$1+y^2 = y^2$

There are no solutions for this, so $x= \pm 1$ is not a solution.

For $x = \pm \dfrac{13}{12} \implies x^2=\dfrac{169}{144}$ , we substitute it into Eq. (3):

$\dfrac{169}{144} + y^2 = \dfrac{169}{144}y^2\\ \dfrac{25}{144}y^2 = \dfrac{169}{144}\\ y^2 = \dfrac{169}{25} \implies y = \pm \dfrac{13}{5}$

We have 4 possible solutions: $\left( \dfrac{13}{12}, \dfrac{13}{5} \right) , \left( -\dfrac{13}{12}, \dfrac{13}{5} \right) , \left( \dfrac{13}{12}, -\dfrac{13}{5} \right) , \left( -\dfrac{13}{12}, -\dfrac{13}{5} \right)$

Substitute all 4 into $f(x,y)$ , and the largest value obtained will be the answer:

$f \left(\dfrac{13}{12}, \dfrac{13}{5} \right) = 20\\ f \left(-\dfrac{13}{12}, \dfrac{13}{5} \right) \approx -2.152 \\ f \left(\dfrac{13}{12}, -\dfrac{13}{5} \right) \approx 16.154 \\ f \left(-\dfrac{13}{12}, -\dfrac{13}{5} \right) = -6$

Therefore, the maximum value of the expression is $\boxed{20}$

We have,

$$

$x^2 + y^2 = x^2y^2$

$$ $\dfrac{1}{x^2} + \dfrac{1}{y^2} = 1 ; x≠ 0, y≠0$

$$ Using the Cauchy-Schwarz inequality,

$\displaystyle \sqrt{a^2 + b^2} \geq \dfrac{ la + mb}{ \sqrt{l^2 + m^2} }$

$$ Putting $a = \dfrac{1}{x}, b = \dfrac{1}{y}, l = 12, m = 5$ , we get

$$ $\sqrt{ \dfrac{1}{x^2} + \dfrac{1}{y^2} }\geq \dfrac{ \dfrac{12}{x} + \dfrac{5}{y} }{13}$

$\sqrt{1} \geq \dfrac{ \dfrac{12}{x} + \dfrac{5}{y} }{13}$

$\dfrac{12}{x} + \dfrac{5}{y} \leq 13$

Hence, $\dfrac{ 5x + 12y + 7xy }{xy} = \dfrac{5}{y} + \dfrac{12}{x} + 7 \leq 13 + 7 \leq 20$

$$

$Thus,$ $the$ $max$ $value$ $of$ $the$ $expression$ $is$ $20.$

-1000000 1 -1 18.999995 -5.000005

2.59 1.084061642 -1.084061642 19.99998317 -2.138979305

2.6 1.083333333 -1.083333333 20 -2.153846154

2.61 1.082614821 -1.082614821 19.99998345 -2.168565821

Excel's answer: 20

The calculus solution can be simplified by making the substitutions a = 1/y, b = 1/x. Then b = sqrt(1-a*a) and E can be expressed in terms of a alone. Differentiation shows that E has a maximum of 20 when a = 5/13.