max. value

[lampat]

If \(\displaystyle x\;,y>0\) and \(\displaystyle x+y = 1\). Then Maximum value of \(\displaystyle x^4y+xy^4 = \)

 

[tkhunny]

Several ways to go about it. Whay material are you studying? What section are you in?

Please show your work so we can have a clue how to help you.

 

[lampat]

I have solved using \(\displaystyle \bf{A.M \geq G.M}\) , Then \(\displaystyle \displaystyle x+y \geq \sqrt{xy}\leftrightarrow xy \leq \frac{1}{4}\)

Now Let \(\displaystyle z = x^4y+xy^4 = xy.\left(x+y \right).\left(x^2+y^2 -xy \right)\)

\(\displaystyle \displaystyle z = \left(xy\right).\left(1-3xy\right) = -3\left\{\left(xy\right)^2-\frac{1}{3}\left(xy\right)+\frac{1}{6^2}-\frac{1}{6^2}\right\}\)

So \(\displaystyle \displaystyle z = \frac{1}{12} - 3\left(xy-\frac{1}{6}\right)^2\)

So \(\displaystyle \displaystyle z_{Max} = \frac{1}{12}-\frac{1}{48} = \frac{3}{48} = \frac{1}{16}\) at \(\displaystyle \displaystyle xy = \frac{1}{4}\)

 

[DrPhil]

I have solved using \(\displaystyle \bf{A.M \geq G.M}\) , Then \(\displaystyle \displaystyle x+y \geq \sqrt{xy}\leftrightarrow xy \leq \frac{1}{4}\)

Now Let \(\displaystyle z = x^4y+xy^4 = xy.\left(x+y \right).\left(x^2+y^2 -xy \right)\)

\(\displaystyle \displaystyle z = \left(xy\right).\left(1-3xy\right) = -3\left\{\left(xy\right)^2-\frac{1}{3}\left(xy\right)+\frac{1}{6^2}-\frac{1}{6^2}\right\}\)

So \(\displaystyle \displaystyle z = \frac{1}{12} - 3\left(xy-\frac{1}{6}\right)^2\)

So \(\displaystyle \displaystyle z_{Max} = \frac{1}{12}-\frac{1}{48} = \frac{3}{48} = \frac{1}{16}\) at \(\displaystyle \displaystyle xy = \frac{1}{4}\)

It would have been easier if you had put in all the steps, but I think I can follow up the the equation

\(\displaystyle z = (xy)(1 - 3xy) \)

At that point I would substitute \(\displaystyle u = xy\), and maximize z with respect to u:

\(\displaystyle \displaystyle z = u - 3u^2 \)

\(\displaystyle \displaystyle \frac{dz}{du} = 1 - 6u = 0 \implies u = \frac{1}{6}\)

\(\displaystyle \displaystyle z_{max} = \frac{1}{6}\left[1 - 3\left(\frac{1}{6}\right) \right] = \frac{1}{12}\)

Your original inequality gives an upper limit for \(\displaystyle xy\), but setting the derivative to 0 gave a smaller value.

 

[DrPhil]

\(\displaystyle \displaystyle z_{Max} = \frac{1}{12}-\frac{1}{48} = \frac{3}{48} = \frac{1}{16}\)

\(\displaystyle \displaystyle z_{max} = \frac{1}{6}\left[1 - 3\left(\frac{1}{6}\right) \right] = \frac{1}{12}\).

Subhotosh Kahn suggested explaining how we came to two different answers.Your approach found the maximum possible value for \(\displaystyle (xy) \le 1/4\), and you then found the corresponding \(\displaystyle z= 1/16\). That is a lower bound for what \(\displaystyle z_{max}\) may be.

When I set the derivative to zero, I found the unique critical value \(\displaystyle (xy) = 1/6\) (which is less than the maximum allowed). Evaluating \(\displaystyle z\) at that value of \(\displaystyle (xy)\) gave \(\displaystyle z_{max} = 1/12\), which is the maximum \(\displaystyle z\) anywhere.

The real "trick" in this problem was all the algebra leading to finding \(\displaystyle z\) as a function of the single variable \(\displaystyle (xy)\). You were way ahead of me doing that!