An algebra problem by Calvin Lin

Let $x=R\cos \theta$ and $y=R\sin \theta$

As both $x>0$ and $y>0$ we shall assume that $R>0$ and $0<\theta<\displaystyle\frac{\pi}{2}$

Substituting the above values in the given relation:

$\displaystyle\frac{1}{R\cos \theta}+\displaystyle\frac{8}{R\sin \theta}=1$

Thus $R=\displaystyle\frac{\sin \theta+8\cos \theta}{\sin \theta\cos \theta}$

So substituting the above values in the expression whose minimum value has to be found we get:

$R\left( \sin \theta+\cos \theta +1\right)$

$=\displaystyle\frac{\sin \theta+8\cos \theta}{\sin \theta\cos \theta}\left( \sin \theta+\cos \theta +1\right)$

Now let the value of the given expression be $f(\theta)$ . Expanding and rearranging we get:

$f(\theta)=9+\displaystyle\frac{1+\sin \theta}{\cos \theta}+8\displaystyle\frac{1+\cos \theta}{\sin \theta}$

Converting to half angle:

$f(\theta)=9+\displaystyle\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}+\displaystyle\frac{8}{\tan \frac{\theta}{2}}$

So $f(\theta)=8+2\left( \displaystyle\frac{1}{1-\tan \frac{\theta}{2}}+\displaystyle\frac{4}{\tan \frac{\theta}{2}}\right)$

Now again we can put $\tan \frac{\theta}{2}=t$ to obtain:

$8+2\left( \displaystyle\frac{1}{1-t}+\displaystyle\frac{4}{t}\right)$

Differentiating w.r.t. $t$ and equating it to $0$ we get the following equation:

$3t^{2}-8t+4=0$

whose roots are $t=\displaystyle\frac{2}{3}$ and $t=2$

Hence $\tan \frac{\theta}{2}=\displaystyle\frac{2}{3}$ or $\tan \frac{\theta}{2}=2$

However note that $\tan \frac{\theta}{2}=2$ is rejected because if it were true then $\theta$ would be greater than $\displaystyle\frac{\pi}{2}$ .

Hence the minimum value must occur at $\tan \frac{\theta}{2}=\displaystyle\frac{2}{3}$

Substituting we obtain $\boxed{26.00}$

This is not a complete solution. For the complete generalized solution, see Unravelling an Inequality Problem .

Consider the geometric interpretation of $x + y + \sqrt{x^2 + y^2}$ . The natural one is to set up a right angled triangle with legs of $x$ and $y$ and a hypotenuse of $\sqrt{x^2+y^2}$ . We are thus after the perimeter of this triangle.

Let $O = (0,0)$ , $A = (x, 0)$ and $B = (0, y )$ . Then, the given condition is equivalent to saying that the point $P=(1, 8)$ lies on the line segment $AB$ . This characterizes all such triangles.

Consider the circle with center $(13, 13)$ and radius 13. This circle is tangential to the x-axis, the y-axis and passes through the point $(1, 8)$ .

Claim: The perimeter is minimized when $AB$ is tangential to this circle at $P$ . Proof: See Unravelling an Inequality Problem .

In particular, the minimum is achieved when $x = \frac{13}{3}$ and $y = \frac{ 52}{5}$ . The value is $2 \times 13 = 26$ .

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How would you generalize this solution? Where did 13 come from?

should the answer be exactly 26.00? i reasoned by thinking that if i minimize (x+y) then i will be able to automatically minimize sqrt(x^2 + y^2). thus, expressing y in terms of x then replace y in (x+y) with it, we get (x^2 - 2x -7)/(1-x). we take the minima which equals 1+2sqrt(2). then solving for y, we get 8+2sqrt(2). finally, substituting the values of x and y in x + y + sqrt(x^2 + y^2) = 26.14...

Please correct me if I'm wrong, I'm getting answer as 26.14 Sorry for not using formats, I'm unable to do so properly.

1/x + 8/y = 1

8x + y = xy

(x -1)(y - 8) - 8 = 0

(x - 1)(y - 8) = 8

(m)(n) = 8 (Let m, n)

For lowest sum value, m and n should be equal to square root. So x = 2sqrt{2} + 1; y = 2sqrt{2} + 8

Putting the values, we get 12 + 10sqrt2 = 26.14