Question

Given that $x^2+y^2=14x+6y+6$, find the largest possible value of the expression $E=3x+4y$.

Answer 1

$x^2-14x+y^2-6y=6$

Adding $49$ and $9$ on both side we get

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

$(x-7{)}^2+(y-3{)}^2=64=8^2$

This is a circle of center $(7,3)$ radius $8$

So,

$y=3+8\sin \theta$

$x=7+8\cos \theta$

Now putting these values in $E=3x+4y$

We have,

$E=3(7+8\cos \theta )+4(3+8\sin \theta )$

$E=21+24\cos \theta )+12+32\sin \theta$

$E=33+24\cos \theta +32\sin \theta ............\left(i\right)$

For maximum value differentiate it with respect to $\theta$

$\frac{dE}{d\theta }=-24\sin \theta +32\cos \theta =0$

$24\sin \theta =32\cos \theta$

$\tan \theta =\frac{32}{24}=\frac{4}{3}$

$\theta =\arctan \left(\frac{4}{3}\right)$

Now putting the value of $\theta$ in (i)

$E=33+24\cos \left(\arctan \left(\frac{4}{3}\right)\right)+32\sin \left(\arctan \left(\frac{4}{3}\right)\right)$

$E=33+24\times 0.6+32\times 0.8$

$E=33+14.4+25.6$

$E=73$