Find Maximum Area

[mathdad]

A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?

Note: I know that calculus can be used to find the maximum area but how is this done using algebra?

 

[MarkFL]

Let \(x\) be the length of the one side parallel to the highway, and \(y\) be the lengths of the two sides perpendicular to the highway, where all lengths are in meters. Then we have:

[MATH]A=xy[/MATH]
[MATH]x+2y=2000[/MATH]
Can you proceed?

 

[mathdad]

Let \(x\) be the length of the one side parallel to the highway, and \(y\) be the lengths of the two sides perpendicular to the highway, where all lengths are in meters. Then we have:

[MATH]A=xy[/MATH]
[MATH]x+2y=2000[/MATH]
Can you proceed?

Yes, I can proceed. I will post my complete solution.

 

[mathdad]

Let \(x\) be the length of the one side parallel to the highway, and \(y\) be the lengths of the two sides perpendicular to the highway, where all lengths are in meters. Then we have:

[MATH]A=xy[/MATH]
[MATH]x+2y=2000[/MATH]
Can you proceed?

A = xy

x + 2y = 2000

Question:

Do I solve x + 2y = 2000 for x or y? I believe it does not matter. I must solve for x or y in order to plug back into A = xy to find my quadratic equation.

x = 2000 - 2y

Plug into area formula.

A = (2000 - 2y)y

A = 2000 - 2y^2

Find y = -b/2a

y = -(2000)/2(-2)

y = -(2000/-4)

y = - (-500)

y = 500

A = 2000 - 2(500)^2

A = 2000 - 2(250,000)

Let me stop here. The answer cannot be a negative area. Mark, where did I goof? I'm sure it is a silly sign error, which I'm famous for.

 

[MarkFL]

You want:

[MATH]A(y)=2000y-2y^2[/MATH]
Had you left it in factored form:

[MATH]A=2y(1000-y)[/MATH]
You would see the roots are \(y=0\) and \(y=1000\) and so the axis of symmetry must be \(y=500\).

 

[mathdad]

You want:

[MATH]A(y)=2000y-2y^2[/MATH]
Had you left it in factored form:

[MATH]A=2y(1000-y)[/MATH]
You would see the roots are \(y=0\) and \(y=1000\) and so the axis of symmetry must be \(y=500\).

Jason's site is still around getting a bit of traffic.

I forgot to distribute y to 2000.

A(500) = 2000(500) - 2(500)^2

A(500) = 1,000,000 - 2(250,000)

A(500) = 500,000

The largest area is 500,000 (meters)^2.

 

[JeffM]

The answer is correct. The factored version can be used as a cross-check on your own work.

[MATH]A = 2y(1000 - y).[/MATH]
[MATH]\therefore max(A) = 2 * 500 * (1000 - 500) = 1000 * 500 = 500,000.[/MATH]
A more interesting problem is how do you KNOW that y = 500 gives the maximum area WITHOUT calculus. (That the book said so is not a mathematician's answer to that question.) How do you check that?

Suppose you experimented numerically and thus guessed that y = 500 is the correct answer.

[MATH]A(500 + h) = 2(500 + h)\{1000 - (500 + h) = 2(500 + h)(500 - h) =[/MATH]
[MATH]2(250000 - h^2) = 500000 - 2h^2 < 500,000 \text { unless } h = 0.[/MATH]
And this gives an intuitive way to introduce differential calculus.

 

[mathdad]

The answer is correct. The factored version can be used as a cross-check on your own work.

[MATH]A = 2y(1000 - y).[/MATH]
[MATH]\therefore max(A) = 2 * 500 * (1000 - 500) = 1000 * 500 = 500,000.[/MATH]
A more interesting problem is how do you KNOW that y = 500 gives the maximum area WITHOUT calculus. (That the book said so is not a mathematician's answer to that question.) How do you check that?

Suppose you experimented numerically and thus guessed that y = 500 is the correct answer.

[MATH]A(500 + h) = 2(500 + h)\{1000 - (500 + h) = 2(500 + h)(500 - h) =[/MATH]
[MATH]2(250000 - h^2) = 500000 - 2h^2 < 500,000 \text { unless } h = 0.[/MATH]
And this gives an intuitive way to introduce differential calculus.

I look forward to calculus 1 after my college algebra review.

 

[Steven G]

I forgot to distribute y to 2000.

A(500) =2000(500) - 2(500)^2

A(500) = 1,000,000 - 2(250,000)

A(500) = 500,000

The largest area is 500,000 (meters)^2.

i would factor out the 500.
2000(500) - 2(500)^2 = 500(2000-2*500) = 500(2000 - 1000) = 500(1000) = 500,000