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 [tex]y= \frac{16-4x}{6} [/tex].
2.
[tex]A_{square}=x^2 \\ A_{rectangle}=y\cdot 2y=\frac{16-4x}{6}\cdot2\cdot\frac{16-4x}{6}= \frac{(16-4x)^2}{18} \\ A(x)=A_{square}+A_{rectangle}=x^2+\frac{(16-4x)^2}{18}[/tex].
3. Find the derivative of the function A(x):
[tex]A'(x)=2x+ \dfrac{2(16-4x)\cdot(-4)}{18} =2x- \dfrac{4(16-4x )}{9} [/tex].
4. Solve the equation A'(x)=0:
[tex]2x- \dfrac{4(16-4x )}{9}=0 \\ 18x-64+16x=0 \\ 34x=64 \\ x= \frac{32}{17} [/tex]
5. Since [tex]y= \frac{16-4x}{6} [/tex] you have
[tex]y= \dfrac{16-4 \frac{32}{17} }{6} = \dfrac{16\cdot17-4\cdot32}{6\cdot 17} = \dfrac{24}{17} [/tex].
Answer: the width of the rectangle is [tex] \frac{24}{17} [/tex]
