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?

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- Mar 3rd 2010, 01:19 PMdeltaXMinimum
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?

- Mar 3rd 2010, 01:34 PMArchie Meade
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

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

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

$\displaystyle 1152+2y^2=ky$

Differentiate and equate to zero to find minimum y

$\displaystyle 4y=k$

$\displaystyle 1152=2y^2$

$\displaystyle y=\sqrt{576}=24$

$\displaystyle x=\frac{384}{24}=16$ - Mar 3rd 2010, 02:55 PMdeltaX
- Mar 3rd 2010, 03:04 PMArchie Meade
Hi deltax,

sorry, that was a few steps in one go!

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

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

multiplying by y

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

Differentiate

$\displaystyle 4y=k$

Now, use this instead of k in (1)

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

Subtract $\displaystyle 2y^2$ from both sides

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

$\displaystyle 1152=2y^2$

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

$\displaystyle y=\sqrt{576}$