Maximizing Area with Limited Fencing: Finding the Perfect Dimensions

[Vigo]

______y__________y______
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l--x---------l--x-----------l x
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A person has 400 ft. of fencing to maximize the area. What are the dimensions?

400=4y + 3x
A=2xy

57.143=y+x
y=57.143-x

A=2(57.143-x)(x)
A=114.286x-2x^2
A'=114.286-4x=0
114.286=4x
x=28.572

400=4y+3(28.572)
400=4y+85.715
315.286=4y
y=78.571

x=28.572 ft.
y=78.571 ft.

Does this look right?

 

[benorin]

$$400=4y + 3x \Rightarrow y=\frac{400-3x}{4}$$

substitute that into the area formula to get

$$A(x)=2xy= 200x-\frac{3}{2}x^2$$

hence

$$A^{\prime}(x)=\frac{d}{dx}\left( 200x-\frac{3}{2}x^2\right) = 200-3x=0\Rightarrow x= \frac{200}{3}\approx 66.67$$

and recall that $y=\frac{400-3x}{4}$, so

$$y=\frac{400-3\frac{200}{3}}{4}= 50$$

so the dimensions are (roughly) 66.67 by 50.

 

[Vigo]

Thank you very much for correcting me. I guess my error was solving for y. I don't know how I made that mistake but thanks a lot.