SOLUTION: An 8 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 wi

Question 1135912: An 8 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.
What is the width of the rectangle that will minimize the total area?
Answer by ikleyn(52800) (Show Source): You can put this solution on YOUR website!
.

Let x = the width of the rectangle (now unknown).


Then its length is 2x, according to the condition.


Then the perimeter of the rectangle is  x + 2x + x + 2x = 6x.


Then the perimeter of the square is the rest  8-6x  and its side is   = .


Then the total area is   = x*(2x) + .


Simplify this expression;  then take the derivative over x and equate it to zero.


In this way you will find x.


The rest is just technique.

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SOLUTION: An 8 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 wi

SOLUTION: An 8 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 wi