Question

a 16 inch wire is to be cut. one piece is to be bent into the shape of a​ square, whereas the other piece is to be bent into the shape of a rectangle whose length is twice the width. find the width of the rectangle that will minimize the total area.

Answer 1

1. Let x be a length of square side, then the square perimeter is 4x (you need to cut 4x from 16 in. to make a square with side x in.). Then 16-4x is remained length of wire and you have to make form this piece a rectangle with sides one of which is twice bigger than another. If y is length of the smaller side, then 2y is length of the bigger side and rectangle perimeter is y+2y+y+2y=6y. You have 16-4x in. of wire left, so 6y=16-4x and
`y= (16-4x)/(6)`.

2.

`A_(square)=x^2 \\ A_(rectangle)=y\cdot 2y=(16-4x)/(6)\cdot2\cdot(16-4x)/(6)= ((16-4x)^2)/(18) \\ A(x)=A_(square)+A_(rectangle)=x^2+((16-4x)^2)/(18)`.

3. Find the derivative of the function A(x):

`A'(x)=2x+ (2(16-4x)\cdot(-4))/(18) =2x- (4(16-4x ))/(9)`.

4. Solve the equation A'(x)=0:

`2x- (4(16-4x ))/(9)=0 \\ 18x-64+16x=0 \\ 34x=64 \\ x= (32)/(17)`

5. Since
`y= (16-4x)/(6)` you have


`y= (16-4 (32)/(17) )/(6) = (16\cdot17-4\cdot32)/(6\cdot 17) = (24)/(17)`.

Answer: the width of the rectangle is
`(24)/(17)`