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Maximize the area

  • Thread starter fk378
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1. 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?

3. 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.
 

berkeman

Mentor
55,338
5,515
1. 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?

3. 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?
 
Last edited:
367
0
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...?
 

berkeman

Mentor
55,338
5,515
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.

[tex]\frac{d}{dy} (xy - 2y^2) = ?[/tex]
 
Last edited:

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