Maximize Area of Rectangle w/ x Feet Fencing

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Homework Statement


A total of x feet of fencing is to form 3 sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?

The Attempt at a Solution


So far I have 2 sides of the yard is denoted by y, and one side is denoted by z.
Then, 2y+z=x
We want to maximize the Area=yz

Now, do I substitute one of the variables in the Area equation to get the answer? Then I know I need to differentiate, but with respect to what? I'm confused here.
 
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fk378 said:

Homework Statement


A total of x feet of fencing is to form 3 sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?

The Attempt at a Solution


So far I have 2 sides of the yard is denoted by y, and one side is denoted by z.
Then, 2y+z=x
We want to maximize the Area=yz

Now, do I substitute one of the variables in the Area equation to get the answer? Then I know I need to differentiate, but with respect to what? I'm confused here.

You're on the right track. A=yz=y(substitute what for z?)

Then you will have an equation for A in terms of y (the variable) and x (the constant). Since you can vary y to vary the area A, you will differentiate A with respect to y, and then do what to find the value of y that gives you maximum A?
 
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Ok, so I have
A=yz=y(x-2y)=xy-2y^2
A'=y-4y

We want to find the critical points so set A'=0
A'=y-4y=0
y=4y

...that doesn't make sense...?
 
fk378 said:
Ok, so I have
A=yz=y(x-2y)=xy-2y^2
A'=y-4y

We want to find the critical points so set A'=0
A'=y-4y=0
y=4y

...that doesn't make sense...?

Your calculation of A' in the 2nd equation has one incorrect term in it. Remember what you are differentiating with respect to.

\frac{d}{dy} (xy - 2y^2) = ?
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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