Unbalanced Maximum

Rearranging the given equation, we get,

$\left(x-7\right)^2 + \left(y-3\right)^2 = 64$

Hence, we can substitute,

$x = 7 + 8\cos\theta$
$y = 3 + 8\sin\theta$

We are required to evaluate the maximum value of $3x + 4y$

Thus, we find the maximum of $33 + 24\cos\theta + 32\sin\theta$ ,

which is clearly $33 + \sqrt{24^2 + 32^2} = \boxed{73}$

We see square and linear terms, let's try to form perfect squares with them:

$(x^2-14x+49)+(y^2-6y+9)=6+49+9=64$

$(x-7)^2+(y-3)^2=64$

Here we have a sum of squares, which seems suitable for Cauchy's inequality!

Since we are maximizing $3x+4y$ , our multiplier would be $(3^2+4^2)$ :

$64(3^2+4^2)=[(x-7)^2+(y-3)^2](3^2+4^2)\ge [3(x-7)+4(y-3)]^2=(3x+4y-33)^2$

$40^2\ge (3x+4y-33)^2$

$40\ge 3x+4y-33$ (we exclude the other case because that doesn't give us the maximum)

$3x+4y\le 40+33=\boxed{73}$ where equality holds when $y=10, x=11$

From the condition we get $y = \sqrt{-x^2+ 4x + 15} + 3$ (we ignore the second root, because $3x + 4y$ is increasing in $y$ and this root is clearly the larger one). Thus we wish to maximize $f(x) = 3x + 4\left(\sqrt{-x^2+ 4x + 15} + 3\right)$ Note that $f'(x) =\frac{3\sqrt{-x^2 + 14x + 15} - 4x + 28}{\sqrt{-x^2 + 14x + 15}} = 0$ i.e. $3\sqrt{-x^2 + 14x + 15} = 4x - 28$ Squaring both sides yields $(5 x-11) (5 x-59) = 0$ The solution $\frac{11}{5}$ is extraneous, hence $f\left(\frac{59}{5}\right) = \boxed{73}$ is the maximum, which can be also verified by checking convexity/concavity of $f(x)$ .

1) $x^{2}+y^{2}=14x+6y+6$ could be writen as : $(x-7)^{2}+(y-3)^{2}=64$ , this is a cercle with radi $8$ centered at $(7;3)$ .

2) Let $3x+4y=b$ , we search to maximise the parameter $b$ , so let's use this equation in the following form : $y=-\frac{3}{4}x+\frac{b}{4}$ , this represents a line in the $(x;y)$ coordinates.

3) Find where the circle and the line intersect in one point (when the line is tangent to the cercle, two solutions here)

4) We get two values for $b$ : $b=-7$ and $b=73$ .

Therefore the answer is $73$ .

P.S. we could always resort to "Lagrange multipliers" to solve these kind of problems.

One way to solve this is through a clever use of Cauchy-Schwarz Inequality.

First complete the squares to get $(x - 7)^2 + (y - 3)^2 = 64$ .

Our left hand side looks like the "greater than" side of the cauchy inequality. Since we want $3x + 4y$ , multiply by $3^2 + 4^2$ to get $64 = ((x - 7)^2 + (y - 3)^2)(3^2 + 4^2) \geq (3(x - 7) + 4(y - 3))^2$

$64(3^2 + 4^2) \geq (3x + 4y - 33)^2$

$8^2 * 5^2 \geq (3x + 4y - 33)^2$

$40 \geq 3x + 4y - 33$

$3x + 4y \leq 73$

With equality when the terms are linearly dependent. Namely, $\frac{x-7}{3} = \frac{y-3}{4}$ . Using the original equation, equality is reached at $(\frac{59}{5}, \frac{47}{5})$ .

$x^2+y^2=14x+6y+6 \Rightarrow (x-7)^2+(y-3)^2=64 \mbox{. Let : } \\ x'=x-7 \\ y'=y-3 \Rightarrow x'^2+y'^2=64 \\ \mbox{We can find unique numbers } r>0 \mbox{ and } 0\le\theta<360 \mbox{ for which :} \\ x'=rcos(\theta) \\ y'=rsin(\theta) \Rightarrow x'^2+y'^2 = r^2{cos}^2(\theta)+r^2{sin}^2(\theta)=r^2({cos}^2(\theta)+{sin}^2(\theta))=r^2 \\ r^2=64 \Rightarrow r=8 \mbox{,the only root since r>0 .}$

$\\ 3x+4y = 3(x'+7)+4(y'+3) = 3x'+4y'+33 = 3*8cos(\theta)+4*8sin(\theta)+33 \\ 8(3cos(\theta)+4sin(\theta))+33 \mbox{ We can find r' and }\varphi { :} \\ 3=r'sin(\varphi) \\ 4=r'cos(\varphi) \Rightarrow r'=\sqrt{3^2+4^2}=5 \Rightarrow \\ 8(5sin(\varphi)cos(\theta)+5cos(\varphi)sin(\theta))+33=40(sin(\varphi)cos(\theta)+cos(\varphi)sin(\theta))+33\\ =40sin(\theta+\varphi)+33 \mbox{.The maximum value is reached when }sin(\theta+\varphi)=1 \\ \mbox{ ,and that value is :} \boxed{73}$

Consider the set of lines $3x + 4y = C$ .

Obviously there are infinitely many of these lines that intersect the circle $x^2 + y^2 = 14x + 6y + 6$ .

Simplifying the circle equation, we have $(x -7)^2 + (y - 3)^2 = 64$ .

Now looking back at our set of lines, we see it is simply the set of all lines with slope $-\frac{3}{4}$ .

Since linear functions have their maxima at their extrema, our solution line must be one of the two lines with slope -3/4 that is tangent to our circle. (Visualizing and applying common sense, it is clearly the tangent near the upper right corner of the circle).

Solving, we find the equation of the tangent line is $3x + 4y = 73$ .

step 1) Differentiate the given equation w. r. t. x to get the value of dy/dx in terms of x and y. step 2) Put dy/dx = -3/4. Get the equation 4x - 3y = 19. step 3) Let k be the maximum value of 3x + 4y. Solve this with 4x - 3y = 19. Get x and y in terms of k. x = (3k + 76)/25 and y = (4k - 57)/25 step 4) Pur these values of x and y in the given equation. Get quadratic eqn in k. It is k^2 - 66k - 511 = 0. step 5) Solve this quadratic equation. One value is minimum of 3x + 4y. The other value is the maximum of 3x + 4y. This value is 73. Thus solved.

Changing x^{2} +y^{2} = 14x+6y+6 we get (x-7)^2 + (y-3)^2 = 8^{2} which represent equation of circle with center (7,3) and radius 8

For maximum value of 3x+4y : The line 3x+4y=k should be farthest distance from center and touch the circle.

The line passing through (7,3) and parallel to 3x+4y= k is 3x+4y = 33 The farthest distance from center of circle is 8

(|k-33|)/5 = 8

K= 73

Suppose, the equation $x^2 + y^2 = 14x + 6y + 6$ is a equation of circle.

Then we can rewrite the equation as $(x-7)^2+(y-3)^2=8^2$

Converting this into the parametric form gives us - $x=7+8cos \theta$ and $y=3+8sin \theta$

Now, let $A=3x+4y$ . Plugging the parametric form in this equation gives us - $A=21+24cos \theta+12+32sin \theta=33+24cos \theta+32sin \theta$

Now, on taking the derivative of $A$ w.r.t $\theta$ ,and putting it equal to $0$ , we get $tan \theta = \frac{4}{3}$

On taking the second derivative, we also find that this value $(tan \theta = \frac{4}{3})$ will give us the maximum value of $A$ . Thus, we get maximum of $A=33+24cos \theta+32sin \theta$ at $sin \theta=\frac{4}{5}$ and $cos \theta=\frac{3}{5}$ .

Plugging, these values back in $A$ , we find $A_{max}=33+\frac{72}{5}+\frac{128}{5}=\boxed{73}$

But this is an Algebra problem. Let us solve it using Algebra.

$\begin{aligned} x^2+y^2 & = 14x+6y+6 \\ \Rightarrow x^2 - 14x +49 +y^2 - 6y + 9 & = 6 + 49 + 9 \\ (x-7)^2 +(y-3)^2 & = 64 \end{aligned}$

Using Cauchy-Schwartz inequality, we have:

$\begin{aligned} \left[ 3(x-7) +4(y - 3)\right]^2 & \le \left(3^2 + 4^2 \right) \left[ (x-7)^2 + (y -3)^2 \right] \\ \Rightarrow \left[ 3(x-7) +4(y - 3)\right]^2 & \le (25)(64) \\ \Rightarrow 3(x-7) +4(y - 3) & \le (5)(8) \\ 3x-21 +4y - 12 & \le 40 \\ 3x+4y & \le 40 +21 +12 \\ 3x+4y & \le \boxed{77} \end{aligned}$

Let 3x+4y=a. We have to find the max value for a. This is the equation of a straight line with slope -3/4.

As we keep on increasing 'a', the straight line will go farther from origin.

Now, the equation given is a circle with centre (7,3) and radius 8.

Hence we need that straight line which is a tangent to this circle. So, as the radius line to a tangent is perpendicular to it, the slope of the line passing through (7,3) and our required point is 4/3.

So, we find that point on the circle which lies on the line through (7,3) and slope 4/3.

this is a circle........just write the equation of the tangent to it with a slope of -0.75.......... .....write eqn of tangent in slope form......it will automatically give .3x+4y=73.........it also gives the minimum value as -7

$(x-7)^2+(y-3)^2=64$ $3x+4y=3(x-7)+4(y-3)+21+12$ Use Cauchy-Schwarz Inequality $3x+4y =< sqrt(3^2+4^2)sqrt((x-7)^2+(y-3)^2)+21+12$ $=< 5(8)+21+12 =73$

The slope of the tengent to the circle at point (x,y) will follow the equation: $-\frac{(x-7)}{(y-3)}$ and should be equal to the slope of 3x+4y=C for maximum value of "C" hecne $-\frac{(x-7)}{(y-3)}=-\frac{3}{4}$ hence put the value of $x-7=\frac{3}{4}\times(y-3)$ in equation of circle (which having a center of (7,3)) and find X and Y it will be 59/5 and 47/5 respectively hence the ans is 73

x=7+8cos(t) y=3+8sin(t)

max 3x+4y = 24cost(t) + 32sin(t) + 33

At max tan(t) = 4/3 sin(t) = 4/5 cost(t) = 3/5

Hence max is 73

From the prob. We get $(x-7)^{2}+(y-3)^2=64$ From Cauchy $[(x-7)^{2}+(y-3)^2][9+16]>=[3(x-7)+4(x-3)]^{2}$ Then $(64)(25)>=(3x+4y-33)^{2}$ $40>=3x+4y-33$ $73>=3x+4y$ Answer $\boxed{73}$

We can write the equation as (x-7)^2 + (y-3)^2 = 8^2 so x = 8sinA + 7, y = 8cosA + 3

So 3x + 4y = 8(3sinA + 4cosA) + 33 <= 8x5 + 33 = 73