Thread: Geometrical application of differentiation maximums and minimums

1. Geometrical application of differentiation maximums and minimums

i really need help for these 2 questions
I need to use deriviation on these questions and the simplest answer would be the best ><>< thank you

a piece of wire 6 metres long is cut into two parts, one of which is used to form a square and the other to form a rectangle whose length is three times its width. Find the lengths of the two parts if the sum of the areas is a minimum.

and

A rectangular area is to be fenced and divided into six rectangles by one dividing fence parallel to its length and two fences parrallel to its breadth. If the total length of fencing available is 120m fine the maxium possible area.

2. Originally Posted by chibiusagi
a piece of wire 6 metres long is cut into two parts, one of which is used to form a square and the other to form a rectangle whose length is three times its width. Find the lengths of the two parts if the sum of the areas is a minimum.
Call the length of the two parts of the 6 m wire x and y. Then we know that
$x + y = 6$

I will use the length "x" to form the square, so each side of the square is x/4 long. The area of this square will be
$A_{sq} = \left ( \frac{x}{4} \right ) ^2 = \frac{x^2}{16}$

The length of wire "y" is used to create a rectangle whose length (L) is three times its width (W). So L = 3W and the perimeter of this rectangle is 2(L + W) = 8W = y. So W = y/8 and L = 3y/8. Thus the area of the rectangle is
$A_{rec} = LW = \frac{3y^2}{64}$

So the total area of the shapes is
$A = \frac{x^2}{16} + \frac{3y^2}{64}$

$A = \frac{x^2}{16} + \frac{3(6 - x)^2}{64}$
from the $x + y = 6$ condition.

Now minimize this function.

-Dan

3. thank you however .....

thank you Dan however the problem is if i follow ur way i can't make the first deriviation also even if i am able to deriviate it the answer is wrong

4. Hello, chibiusagi!

A rectangular area is to be fenced and divided into six rectangles
by one dividing fence parallel to its length and two fences parrallel to its breadth.
If the total length of fencing available is 120m fine the maxium possible area.
Code:
    - *-------*-------*-------*
: |       |       |       |
y |       |       |       |
: |       |       |       |
- *-------*-------*-------*
: |       |       |       |
y |       |       |       |
: |       |       |       |
- *-------*-------*-------*
: - x - : - x - : - x - :

The amount of fencing is: . $9x + 8y \:=\:120\quad\Rightarrow\quad y \:=\:\frac{120-9x}{8}$ .[1]

The area is: . $A \:=\3x)(2x)\quad\Rightarrow\quad A \:=\:6xy" alt="A \:=\3x)(2x)\quad\Rightarrow\quad A \:=\:6xy" /> .[2]

Substitute [1] into [2]: . $A \;=\;6x\left(\frac{120-9x}{8}\right) \;=\;\frac{9}{4}(40x-3x^2)$

. . And that is the function we must maximize.

Don't forget the dimensions of the area are: $3x\text{ and }2y.$

5. Originally Posted by chibiusagi
thank you Dan however the problem is if i follow ur way i can't make the first deriviation also even if i am able to deriviate it the answer is wrong
Originally Posted by topsquark
Call the length of the two parts of the 6 m wire x and y. Then we know that
$x + y = 6$

I will use the length "x" to form the square, so each side of the square is x/4 long. The area of this square will be
$A_{sq} = \left ( \frac{x}{4} \right ) ^2 = \frac{x^2}{16}$

The length of wire "y" is used to create a rectangle whose length (L) is three times its width (W). So L = 3W and the perimeter of this rectangle is 2(L + W) = 8W = y. So W = y/8 and L = 3y/8. Thus the area of the rectangle is
$A_{rec} = LW = \frac{3y^2}{64}$

So the total area of the shapes is
$A = \frac{x^2}{16} + \frac{3y^2}{64}$

$A = \frac{x^2}{16} + \frac{3(6 - x)^2}{64}$
from the $x + y = 6$ condition.

Now minimize this function.

-Dan
What's so hard?

To pick up where I left off:
$\frac{dA}{dx} = \frac{2x}{16} + \frac{6(6 - x)(-1)}{64} = 0$

$\frac{x}{8} - \frac{3(6 - x)}{32} = 0$

$4x - 3(6 - x) = 0$

$4x - 18 + 3x = 0$

$7x - 18 = 0$

$7x = 18$

$x = \frac{18}{7}$

So the other length is
$y = 6 - x = 6 - \frac{18}{7} = \frac{42}{7} - \frac{18}{7} = \frac{24}{7}$