Since $(x,y)$ is constrained to be in a circle, $3x+4y=73$. (The lines $3x+4y=c$ and the circle should touch each other and should be as far away as possible from origin. From this, $c$ can be evaluated). The point of touching gives max value. Another method that can give answer is the Lagrange multiplier but I just learnt it yesterday. Its makes the solution long and its not in my syllabus. Also, I don't know anything about gradients (But I am beginning to love math after reading about it) Is there a simpler solution?
If $x^2+y^2=14x+6y+6$, find the maximum value of $3x+4y$
1.6k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
This is not an answer since it uses (for illustration purposes) Lagrange multipliers).
I am sure that you will love Lagrange multipliers ! So, let me give you a taste for your problem.
Considering the function $$F=3x+4y+\lambda (x^2+y^2-14x-6y-6)$$ let us compute the three derivatives $$F'_x=\lambda (2 x-14)+3$$ $$F'_y=\lambda (2 y-6)+4$$ $$F'_\lambda=x^2+y^2-14x-6y-6$$ These derivatives must be equal to $0$. From $F'_x=0$, we can extract $\lambda=-\frac{3}{2 (x-7)}$. Plug this result in the second derivative $$F'_y=\lambda (2 y-6)+4=4-\frac{3 (2 y-6)}{2 (x-7)}=\frac{4 x-3 y-19}{x-7}=0$$ So, now, solve for $y$ and obtain $$y=\frac{4 x-19}{3} $$ Plug this result in $F'_\lambda$, expand and simplify to arrive to $$F'_\lambda=\frac{25 x^2}{9}-\frac{350 x}{9}+\frac{649}{9}=0$$ Solving the quadratic gives two roots $x_1=\frac{11}{5}$, $x_2=\frac{59}{5}$ to which will correspond $y_1=-\frac{17}{5}$, $y_2=\frac{47}{5}$. So $3x_1+4y_1=-7$ and $3x_2+4y_2=73$.
So, you get the minimum and maximum value of the function for the given constraint.
Substitute $u=x-7$, $v=y-3.$ Then the problem becomes: Find the maximum value of $3u +4v$ if $u^2 +v^2 =64.$ Now substitute: $u=\frac{k}{3}$ and $v=\frac{l}{4}$ then our problem becomes: Find the maximum value of $k+l$ if $\frac{k^2}{24^2} +\frac{l^2}{32^2} =1$. But we can substitute $k=24\sin \xi, l=32\cos \xi$ hence $$k+l=24\sin \xi +32\cos \xi \leq \sqrt{8^2 \cdot 25} =40$$ Hence the maximum of $3x +4y =73.$