Maths problem! Solution required. (1 Viewer)

[jamersonx]

So yeah..

A piece of wire is 20cm long. it is cut into two portions, the length of one portion being 'x' cm. Each portion is then bent to form a rectangle in which the length is twice the breadth.
a) show that the sum of the areas of the two rectangles is given by
A = 1/18(2x^2 - 40x + 400)

b) For what value of 'x' is this area a minimum?

 

[Timothy.Siu]

So yeah..

A piece of wire is 20cm long. it is cut into two portions, the length of one portion being 'x' cm. Each portion is then bent to form a rectangle in which the length is twice the breadth.
a) show that the sum of the areas of the two rectangles is given by
A = 1/18(2x^2 - 40x + 400)

b) For what value of 'x' is this area a minimum?

one portion is x, the other one is 20-x,
since the opposite sides are equal, for the xcm one, length+breadth=x/2
l=2b 3b=x/2 b=x/6, l=x/3
area of this rectangle is x^2/18
similarly for the other one, l+b=10-(x/2)
l=2b 3b=10-(x/2) b=(20-x)/6 l=(20-x)/3
area of this rectangle is (20-x)^2/18

therefore the sum of Areas=x^2/18+(20-x)^2/18=1/18(2x^2-40x+400)

b) A'=1/18 . (4x-40)
A'=0 when x=10 then u test if its a minimum,
A''=1/18(4) >0 therefore x=10 is a minimum

 

[AlexJB]

Just think about it logically.

A piece of wire is 20cm long. If you cut it up into two portions, with one portion x, then the other must be 20-x. Both portions must add up to 20cm. x + (20 - x) = 20 = true.

Now let's just handle the x portion. We know that the length is twice the size of the breadth. Draw up a small rectangle, and label the length and breadth appropriately. Now, as we know the length is twice the breadth, on your diagram just rewrite the length variable (l) as 2b (2x breadth). Therefore we should have length = 2b, breadth=b. The perimiter of the rectangle would therefore be 2b+2b+b+b=6b=x. Therefore b = x/6. Area of a rectangle is length x breadth, =



We've found the area of the rectangle with the x portion, now do the same with the 20-x portion.

Length = 2b, Breadth = b.
2b + 2b + b + b = 6b = 20 - x (perimiter of the rectangle)





Sum of the two =



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To find the min, just differentiate that. Should get 1/18(4x - 40).
Find the second derivative = 4/18.
When f'(x) = 0
4x - 40 = 0
x = 10
When x = 10 f''(10) = 4/18 = >0, therefore min point.
When x = 10 f(10) = 200/18?