Maxima and Minima

Calculus Level 5

Subject to $x+y=1$ for positive reals, the maximum and local minimum of $x^4 y + y^4 x$ can be expressed as $\frac{a}{b}$ and $\frac{c}{d}$ , respectively, where $a,b$ and $c,d$ are pairs of positive coprime integers.

Find $a+b+c+d$ .

The answer is 30.

2 solutions

Discussions for this problem are now closed

Tony Sprinkle
Jan 4, 2015

Using the constraint $x + y = 1$ , we can rewrite the expression in terms of $x$ alone:

$f(x) = x^4(1-x) + (1-x)^4x = x - 4x^2 + 6x^3 - 3x^4$

Taking a derivative and setting the result equal to $0$ reveals critical points:

$f'(x) = 1 - 8x + 18x^2 - 12x^3 = 0\quad\Rightarrow\quad x = \frac{1}{2}, \frac{1}{2} \pm \frac{\sqrt{3}}{6}$

Testing the sign of $f'(x)$ near the critical points, we discover that $f$ has a local minimum at $x=\frac{1}{2}$ and a global maximum at $x = \frac{1}{2} \pm \frac{\sqrt{3}}{6}$ .

$f\left(\frac{1}{2}\right) = \frac{1}{16} \\ f\left(\frac{1}{2} \pm \frac{\sqrt{3}}{6}\right) =\frac{1}{12}$

Thus, $a = 1,\: b = 16,\: c = 1,$ and $d = 12,$ and $a + b + c + d = \boxed{30}$ .

Moderator note:

Ideally, you should show that by the second derivative test, the results you've found are indeed extrema points.

Shivam Hinduja
Dec 19, 2014

Notify me if there any other answer or maxima or minima for this expression

Edit: There must have been 3a in place of a. But the maxima is still correct

Points about your answer:

Your derivation for maximum value is correct.
AM-GM is applicable only when the parameters involved are non negative.
Even if you make $x$ and $y$ positive, your argument that minimum value will occur at $xy=\frac{1}{4}$ isn't reasonable. Think about it.

In fact the expression doesn't have any minimum value. All it has is a local minima at $x =\frac{1}{2}$

I've edited your question. Now it's accurate.

Sanjeet Raria - 6 years, 5 months ago

You are right sir. I almost rejected the case when (x,y) is (1,0) or (0,1). Which would have given another value for xy as 0.

Moreover I forgot that a quadratic can has a maxima and a minima at the same time.

Thanks for reviewing my problem.

Shivam Hinduja - 6 years, 5 months ago

How can we neglect (x,y)=(0,1) or (1,0) ??

Jibin Joy K - 6 years, 5 months ago

@Jibin Joy K – Well, I think that no local minima lies at that point. Moreover if that's the argument then there are many more smaller negative values of the expression which corresponds global minima to be negative ∞ . You must see it's graphical representation.