Leave out all the rest!

The equation can be re-written as: $3(x-3)^2+4(y-4)^2=300$ And the expression we have to maximize is: $z=(x+y-7)^2$ The obvious way to simplify this would be to take (x-3)=a , (y-4)=b : $3a^2+4b^2=300$ $z=(a+b)^2$ Notice how the given equation has a linear combination of the squares of a and b individually, while the function we have to maximize has the square of a linear combination of a and b . Also, f(x)=x^2 is a concave function, so we are obviously inclined to use Jensen's Inequality . It is easy to re-phrase the equation to suit our needs with some guesswork, but suitable co-efficients can also be obtained by forming algebraic equations which are fairly easy to solve (I'm omitting that part here). $\frac { \frac { 1 }{ 3 } { (3a) }^{ 2 }+\frac { 1 }{ 4 } { (4b) }^{ 2 } }{ \frac { 1 }{ 3 } +\frac { 1 }{ 4 } } \ge { \left(\frac { \frac { 1 }{ 3 } .3a+\frac { 1 }{ 4 } .4b }{ \frac { 1 }{ 3 } +\frac { 1 }{ 4 } } \right) }^{ 2 }$ $\therefore \frac { 300 }{ \frac { 1 }{ 3 } +\frac { 1 }{ 4 } } \ge { \left(\frac { a+b }{ \frac { 1 }{ 3 } +\frac { 1 }{ 4 } } \right) }^{ 2 }$ $\therefore { (a+b) }^{ 2 }\le 300\left(\frac { 1 }{ 3 } +\frac { 1 }{ 4 } \right)=\boxed{175}$

We can rewrite the equation as:

$3(x-3)^2-27+4(y-4)^2-64+91=300$

$((x-3)\sqrt3)^2+(2(y-4))^2=(10\sqrt3)^2$

Set $(x-3)\sqrt3=10\sqrt3 \cos t$ and hence $2(y-4)=10\sqrt3 \sin t$ . We now need to find the maximum value of:

$x^2+y^2+2xy-14x-14y+49$

$=(x+y)^2-14(x+y)+49$

$=(x+y-7)^2 = (x-3+y-4)^2$

$=(10\cos t+5\sqrt3 \sin t)^2$

$=100\cos^2 t+75\sin^2 t +100\sqrt3 \sin t \cos t$

$=75+25\cos^2 t + 100\sqrt3 \sin t \cos t$

$=75+\frac{25}2(1+\cos 2t)+50\sqrt3 \sin 2t$

$=75+\frac{25}2+\frac{25}2\cos\phi + 50\sqrt3 \sin\phi$ .

This takes the following maximum (see here for details):

$75+\frac{25}2+\sqrt{\left(\frac{25}2\right)^2+\left(50\sqrt3\right)^2}$

$=75+\frac{25}2\left(1+\sqrt{1+48}\right)$

$=75+8\cdot\frac{25}2$

$=\boxed{175}.$

Use Titu's Lemma!

The first equation can be rewritten as $\frac{(x-3)^2}{\frac{1}{3}}+\frac{(y-4)^2}{\frac{1}{4}}=300$

Then, by Titu's Lemma, $\frac{(x-3)^2}{\frac{1}{3}}+\frac{(y-4)^2}{\frac{1}{4}} \geq \frac{(x+y-7)^2}{\frac{1}{3}+\frac{1}{4}}$

Since $\frac{(x-3)^2}{\frac{1}{3}}+\frac{(y-4)^2}{\frac{1}{4}}=300$ ,

$300 \geq \frac{(x+y-7)^2}{\frac{1}{3}+\frac{1}{4}}$

$175 \geq (x+y-7)^2$ , and since this expression is what we were looking for, our answer is 175.

The first equation represents an ellipse whereas the second expression represents a pair of coincident lines...(talkin in 2d terms). To maximize the expression we have to find the point in the first quadrant where the lines act as tangent to the given ellipse.. The equation of the tangent comes out to be x+y=22.228