3 Rectangular Gardens

**magentarita** · Dec 11th 2008, 08:22 PM

A gardener had 1500 feet of fencing to enclose three adjacent rectangular gardens. Determine the demensions that will produce a maximum enclosed area.

**earboth** · Dec 12th 2008, 11:10 AM

Originally Posted by magentarita

A gardener had 1500 feet of fencing to enclose three adjacent rectangular gardens. Determine the demensions that will produce a maximum enclosed area.

The area of the 3 gardens is calculated by:

$a = l\cdot w$ ...... (A)

The fence of all 3 gardens is calculated by:

$f = 2l + 4w = 1500$ ...... (B)

Solve (B) for l. Plug in this term into equation (A):

$a(w)=w\cdot \left( \frac12 f - 2w \right) = -2w^2 + \frac12 f w$

To get the maximum area determine the first derivation of a and solve the equation $a'(w) = 0$ for $w$ . I've got $w = \frac18 f$

That means the maximum area is

$a\left(\frac18 f\right)= -2\left( \frac18 f \right)^2 + \frac12 f \cdot \frac18 f = \frac1{32} f^2$

**magentarita** · Dec 12th 2008, 04:32 PM

Originally Posted by earboth

The area of the 3 gardens is calculated by:

$a = l\cdot w$ ...... (A)

The fence of all 3 gardens is calculated by:

$f = 2l + 4w = 1500$ ...... (B)

Solve (B) for l. Plug in this term into equation (A):

$a(w)=w\cdot \left( \frac12 f - 2w \right) = -2w^2 + \frac12 f w$

To get the maximum area determine the first derivation of a and solve the equation $a'(w) = 0$ for $w$ . I've got $w = \frac18 f$

That means the maximum area is

$a\left(\frac18 f\right)= -2\left( \frac18 f \right)^2 + \frac12 f \cdot \frac18 f = \frac1{32} f^2$

Can this be done without using derivatives?

**earboth** · Dec 12th 2008, 10:18 PM

Originally Posted by magentarita

Can this be done without using derivatives?

Of course. The graph of the function a(w) is a parabola opening down. Thus the maximum occurs at the vertex of the parabola:

$a(w)=-2w^2+\frac12fw = -2\left(w^2-\frac14fw{\color{red}{+\left(\frac18f \right)^2}} \right){\color{red}{+2\left(\frac18f \right)^2}}$

The vertex has the coordinates $V\left(\frac18f\ ,\ \frac1{32}f^2 \right)$

Therefore the area has a maximum of $\frac1{32}f^2$ at $w = \frac18f$

**magentarita** · Dec 14th 2008, 07:52 AM

Originally Posted by earboth

Of course. The graph of the function a(w) is a parabola opening down. Thus the maximum occurs at the vertex of the parabola:

$a(w)=-2w^2+\frac12fw = -2\left(w^2-\frac14fw{\color{red}{+\left(\frac18f \right)^2}} \right){\color{red}{+2\left(\frac18f \right)^2}}$

The vertex has the coordinates $V\left(\frac18f\ ,\ \frac1{32}f^2 \right)$

Therefore the area has a maximum of $\frac1{32}f^2$ at $w = \frac18f$

I thank you for your help.

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