# Finding Maximum Value

#### anemone

##### MHB POTW Director
Staff member
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

##### MHB Oldtimer
Staff member
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 $\frac54.$

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#### MarkFL

Staff member
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

##### MHB POTW Director
Staff member
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 $$\displaystyle \frac{225}{32}$$.