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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.

Respuesta :

frika
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]




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