Maximum value of $(1 − x)(2 − y)^ 2 (x + y)$

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$x < 1, y<2, x+y>0$ I tried doing AM-GM (since all of the terms are positive.

$(1 − x)(2 − y)^ 2 (x + y) = (2-2x)(2-y)(2-y)(2x+2y)*1/4$ this way we can cancel the x and y's

thus the maximum value is

$((2-2x+2-y+2-y+2x+2y+1/4)/5)^5$

but the answer is said to be 1(?) where $x =0,$ and $y = 1$

Can someone tell me what I did wrong.

Answer 1

In your AM-GM the equality occurs for $$2-2x=2-y=2x+2y=\frac{1}{4},$$ which is impossible.

I think, it's better to make the following.

By AM-GM $$(1-x)(2-y)^2(x+y)=4(1-x)\left(1-\frac{y}{2}\right)^2(x+y)\leq$$ $$\leq4\left(\frac{1-x+2\left(1-\frac{y}{2}\right)+x+y}{4}\right)^4=\frac{81}{64}.$$ The equality occurs for $$1-x=1-\frac{y}{2}=x+y$$ or $$(x,y)=\left(\frac{1}{4},\frac{1}{2}\right),$$ which says that we got a maximal value.

Answer 2

A much more Logical way to solve this without algebraic manipulations:

Since we want to maximize $(1-x)(2-y)^2(x+y)$ , and all the terms are positive, we can see that we might want to maximize $(2-y)^2$ in particular as it is the biggest among all terms.

$$\max \space(2-y)^2 = (2-0)^2 = 4 \space\space\space\space\space\space\space \text for \space y = 0$$

So the equation is reduced to $$(1-x)\times 4\times(x+0)$$ $$= 4\times(1-x)\times x$$ Again to $\max$ this equation further , we want to maximize $(1-x)\times x$ , which happens at $x=\frac 12.$

Hence $$\max \space[(1-x)(2-y)^2(x+y)] = \left(1-\frac12\right) \times (2-0)^2 \times \left(\frac12 + 0 \right)$$ $$ = \frac12\times4\times\frac12 = 1$$