Finding Maximum Value with Partial Differentiation

[anemone]

Find the maximum of the expression \(\displaystyle x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4\) if \(\displaystyle x,\;y\) are real numbers with \(\displaystyle x+y=2\).

 

[Opalg]

Find the maximum of the expression \(\displaystyle x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4\) if \(\displaystyle x,\;y\) are real numbers with \(\displaystyle x+y=2\).

This may not be the quickest solution, but it avoids calculus. Let $x=1+t$, then $y=1-t$. Notice that $1+x+x^2+x^3 = \dfrac{x^4-1}{x-1} = \dfrac{(1+t)^4-1}{t}$, and similarly $1+y+y^2+y^3 = -\dfrac{(1-t)^4-1}{t}.$ Also $xy = 1-t^2.$ Then $$ \begin{aligned}x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4 &= xy\bigl((1+x+x^2+x^3) + (1+y+y^2+y^3) - 1\bigr) \\ &= (1-t^2)\Bigl(\frac{(1+t)^4-1}{t} - \frac{(1-t)^4-1}{t} - 1\Bigr) \\ &= (1-t^2)(7+8t^2) \\ &= 7+t^2 -8t^4 \\ &= \frac{225}{32} - 8\Bigl(t^2 - \frac1{16}\Bigr)^2\quad \text{(completing the square).}\end{aligned}$$ Thus the maximum value is $\frac{225}{32}$, which occurs when $t = \pm\frac14$, or when $x = \frac34$ or $\frac54.$

 

[MarkFL]

Here's a method involving the calculus:

If we use the constraint to get \(\displaystyle y=2-x\) and substitute this into the objective function, we find, after simplification that:

\(\displaystyle f(x)=-8x^4+32x^3-47x^2+30x\)

Equating the derivative to zero:

\(\displaystyle f'(x)=-32x^3+96x^2-94x+30=0\)

Dividing through by 2, we have:

\(\displaystyle -16x^3+48x^2-47x+15=0\)

Multiplying through by -1 and factoring, we have:

\(\displaystyle (x-1)(4x-5)(4x-3)=0\)

Use of the first derivative test shows that relative maxima occur at:

\(\displaystyle x=\frac{3}{4},\,\frac{5}{4}\)

and we find:

\(\displaystyle f_{\text{max}}=f\left(\frac{3}{4} \right)=f\left(\frac{5}{4} \right)=\frac{225}{32}\)

 

[anemone]

Thanks to both of you for participating and also the awesome method on how to solve this problem too!

My solution:

\(\displaystyle x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4=xy(x^3+x^2+x+1+y+y^2+y^3)\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=xy((x^3+y^3)+(x^2+y^2)+(x+y)+1)\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=xy((x+y)^3-3xy(x+y)+(x+y)^2-2xy+(x+y)+1)\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=xy((2)^3-3xy(2)+(2)^2-2xy+(2)+1)\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=xy(15-8xy)\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=-8(xy-\frac{15}{16})^2+\frac{225}{32}\)

Hence, the maximum value is \(\displaystyle \frac{225}{32}\).

 

[CellCoree]

To find the maximum value of this expression, we can use the method of partial differentiation. Taking the partial derivative with respect to x and setting it equal to 0, we get:

4x^3y + 3x^2y + 2xy + y + y^2 + y^3 + y^4 = 0

Rearranging, we get:

x^3y + x^2y + xy + y + y^2 + y^3 + y^4 = 0

Factoring out y, we get:

y(x^3 + x^2 + x + 1 + y + y^2 + y^3) = 0

Since x and y are real numbers, we can set y = 0 to get the maximum value of the expression. Plugging in y = 0, we get:

x^4 + x^3 + x^2 = 0

Solving for x, we get x = 0 or x = -1. Substituting these values back into the original expression, we get the maximum value to be 0.

Therefore, the maximum value of the expression x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4 is 0, when x = 0 and y = 2, or when x = -1 and y = 3. This means that the maximum value occurs when the real numbers x and y are either 0 and 2, or -1 and 3, and their sum is 2.