Question

there are 12 sticks of 1-1 metre make a rectangle using all these sticks whose area is greatest​

Answer 1

Answer:

$\huge \orange{9 \: {m}^{2} }$

Step-by-step explanation:

Let the sides of rectangle be x and y . Now total length of sticks will be parameter of rectangle . so

$2(x + y) = 12 \\ \\ x + y = 6 \\ \\ y = 6 - x$

Now are of rectangle will be

$A= xy = x(6 - x) =6x - {x}^{2}$

Now since sticks can't be broken so we will look for positive integer solutions of x .

Now either you can put x = 1 , 2 , 3 , 4 and 5 or you can think as symmetry that area will be maximum if it's square i.e. x = y = 3 .

Or if you know calculus then for maxima

$\frac{dA}{dx} = 0 \\ \\ \frac{d}{dx} (6x - {x}^{2} ) \\ \\ 6 - 2x = 0 \\ \\ x = 3$

And area

$A = 6x - {x}^{2} = 6 \times 3 - {3}^{2} = 9$

Answer 2

Given :  there are 12 sticks of 1-1 meter make a rectangle

To Find :  using all these sticks whose area is greatest​

Solution:

Stick on one side  = x

Stick on other side = y

2(x + y)  = 12

=> x + y  = 6

Area = xy  

x            y          A=xy

1             5            5

2            4            8

3            3           9

4            2            8       ( different orientation of x = 2 and y = 4)

5             1           5      ( different orientation of x =1 and y = 5)

In all the possible rectangles

Greatest area is  9 m²

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