Max and Min help needed ASAP (1 Viewer)

[Petinga]

A piece of wire of length 50cm is cut into 2 sections. One section is used to construct a rectangle whose dimensions are in the ratio 1:3; the other section is used to construct a square. Find the dimensions of the rectangle and the square so that the total enclosed area is minimum.


Help needed ASAP

 

[rama_v]

A piece of wire of length 50cm is cut into 2 sections. One section is used to construct a rectangle whose dimensions are in the ratio 1:3; the other section is used to construct a square. Find the dimensions of the rectangle and the square so that the total enclosed area is minimum.


Help needed ASAP

Let one length of the wire be x cm and the other be (50-x)cm . So the rectangles dimensions are x(3x)=3x² and the square's dimensions are (50-x)².

I'm not sure if they want the added areas to be a minimum although I think this is how you would do it:
Area = 3x² + (2500 - 100x + x²)
dA/dx = 6x -100 +2x = 8x-100
8x-100 = 0
8x=100, x = 100/8 = 12.5
d²x/dx² = 8 .: minimum
so sub in x = 12.5 back into the areas of the square and the rectangle and you get:
Square: Side length 37.5 cm
Rectangle: 12.5cm x 37.5cm

 

[PtolemysPenguin]

A piece of wire of length 50cm is cut into 2 sections. One section is used to construct a rectangle whose dimensions are in the ratio 1:3; the other section is used to construct a square. Find the dimensions of the rectangle and the square so that the total enclosed area is minimum.


Help needed ASAP

(Note: My x is differant to rama_v's x, sorry for any confusion)

I interpreted this question somewhat differently to rama_v, I think it means that the wire is used to make all of the rectangle and square, not just the bases.

Let the piece of wire used to make the square be x cm, and the other piece be 50 - x cm.

Therefore the square has dimensions x/4 * x/4, and the rectangle has dimensions (50 - x)/8 * 3(50 - x)/8

Therefore the sum of the areas is f(x) = (x^2)/16 + (3/64)(2500 - 100x + x^2)

f'(x) = (14/64)x - 300/64

When f'(x) = 0, then (14/64)x = 300/64, therefore x = 300/14

f''(x) = 14/64, which is positive, therefore x = 300/14 is a minimum

Therefore the minimum area has x = 300/14 = 21 6/14 cm

Therefore the dimensions of the square are 75/14 cm * 75/14 cm, and the dimensions of the rectangle are 75/28 cm * 225/28 cm