# Minimum

• March 3rd 2010, 01:19 PM
deltaX
Minimum
A 384 - square meter plot of land is to be enclosed by a fence and divide into two equal parts by another fence parallel to one pair of sides. What dimensions of the outer rectangle will minimize the amount of fence used?
• March 3rd 2010, 01:34 PM
Quote:

Originally Posted by deltaX
A 384 - square meter plot of land is to be enclosed by a fence and divide into two equal parts by another fence parallel to one pair of sides. What dimensions of the outer rectangle will minimize the amount of fence used?

Take the sides to which the inner fence is parallel to as having length x.
Then the other two sides have length y.

Fence length = 3x+2y

Area =xy=384

Substitute x or y into the linear equation

$xy=384\ \Rightarrow\ x=\frac{384}{y}$

$\frac{3(384)}{y}+2y=k$

$1152+2y^2=ky$

Differentiate and equate to zero to find minimum y

$4y=k$

$1152=2y^2$

$y=\sqrt{576}=24$

$x=\frac{384}{24}=16$
• March 3rd 2010, 02:55 PM
deltaX
Quote:

Take the sides to which the inner fence is parallel to as having length x.
Then the other two sides have length y.

Fence length = 3x+2y

Area =xy=384

Substitute x or y into the linear equation

$xy=384\ \Rightarrow\ x=\frac{384}{y}$

$\frac{3(384)}{y}+2y=k$

$1152+2y^2=ky$

Differentiate and equate to zero to find minimum y

$4y=k$

$1152=2y^2$

$y=\sqrt{576}=24$

$x=\frac{384}{24}=16$

I was just somehow confused on one equation that is: $1152=2y^2$

where did that come from?
• March 3rd 2010, 03:04 PM
Hi deltax,

sorry, that was a few steps in one go!

$\frac{3(384)}{y}+2y=k$

$\frac{1152}{y}+2y=k$

multiplying by y

$1152+2y^2=ky$ (1)

Differentiate

$4y=k$

Now, use this instead of k in (1)

$1152+2y^2=4y(y)=4y^2$

Subtract $2y^2$ from both sides

$1152+2y^2-2y^2=4y^2-2y^2$

$1152=2y^2$

$y^2=\frac{1152}{2}=576$

$y=\sqrt{576}$