Thread: 3 Rectangular Gardens

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

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:

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

The fence of all 3 gardens is calculated by:

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

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

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

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

That means the maximum area is

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

Originally Posted by earboth

The area of the 3 gardens is calculated by:

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

The fence of all 3 gardens is calculated by:

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

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

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

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

That means the maximum area is

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

Can this be done without using derivatives?

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:

$\displaystyle 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 $\displaystyle V\left(\frac18f\ ,\ \frac1{32}f^2 \right) $

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

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:

$\displaystyle 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 $\displaystyle V\left(\frac18f\ ,\ \frac1{32}f^2 \right) $

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

I thank you for your help.