How to find an area? I have this question:
A farmer plans to enclose a rectangular pasture adjacent to a river. The pasture must contain $320,000$ square meters in order to provide enough grass for the herd. What dimensions will require the least amount of fencing if no fencing is needed along the river? 
$$x =    m$$
$$y =    m$$
That what I did not sure what I'm doing wrong?
$$xy=320000$$
$$2(x+y)=P$$ 
$$x+\frac{320,000}{x}=\frac{P}{Z}$$
$$1-\frac{320,000}{x^2}=0$$
$$x=400$$
$$y=400$$
My answer is wrong. I think $X$ needs to be larger then $Y$.
 A: The length of the fence (which is not the whole perimeter) is $2x+y$ (assuming I have the orientation of the axes right). Note that $xy$ is fixed and you want to minimise $2x+y$. We note that  $$(2x+y)^2=(2x-y)^2+8xy$$
With $xy$ fixed, the left hand side of this expression is a minimum when $2x=y$.
A: You set up the problem correctly, though we want for $2x + y = P$
$$ 2(x+y)= 2x + y = P$$
$$P = 2x+\dfrac{320,000}{x} = \frac{2x^2 + 320000}{x}$$
We want to minimize the perimeter needed, so we need to find the derivative of $P(x)$ and set that equal to zero, then solve for the zeros: the $x$ values that solve $P'(x) = 0$. Using the quotient rule: 
$$P'(x) = \dfrac{4x\cdot x - (2x^2 + 320000)}{x^2} $$
Simplify and set equal to zero:
$$P'(x) = \dfrac{2x^2 - 320000}{x^2} = 0$$
$P'(x)$ will equal zero when the numerator is equal to zero. We certainly don't want a rectangle where $x = 0$, so we can affirm $x \neq 0 $ and hence the denominator will not be zero.
So, we simply solve for $x$: $$2x^2 - 320000 = 0 \iff x^2 - 160000 = (x + 400)(x - 400) = 0$$
So $$x = 400, y = \dfrac{320000}{400} = 800.$$
A: Another way to solve without calculus:
For a fence-length $P$, area $A$, and dimensions $x$ and $y$ ($y$ parallel to river):$$x\times y = A$$ $$2x+y=P$$ Re-arranging the second equation to eliminate $y$ from the first:$$x\times (P-2x)=A$$ This re-arranges to the quadratic$$2x^2-Px+A=0 $$ with solutions:$$\frac{P ± \sqrt{ P^2-8A} }{4}$$The smallest value of P that yields a real root is $$P=\sqrt{8A}$$ and thus $$x=\frac{\sqrt{8A}}{4}=400$$
