Question 549580: A piece of wire 16 cm long is cut into two pieces, with the first piece having length x. The first piece is formed into a rectangle in which the length is twice the width. The second piece of wire is also formed into a rectangle, but with the length three times the width. For what value of x is the total area of the two rectangles a minimum?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Since  is the perimeter of the first rectangle where the length is twice the width, we can say:
From which we can derive:
Likewise, since the perimeter of the other rectangle is  and the length of the second rectangle is 3 times the width, we can say:
From which we can derive:
The area of the first rectangle in terms of its width:
And the area of the second rectangle in terms of its width is:
Then the total area is given by:
Substituting the earlier derived expressions for  and  :
\ =\ 2\left(\frac{x}{6}\right)^2\ +\ 3\left(\frac{16\ -\ x}{8}\right)^2)
If you expand, collect terms, and find a common denominator, you will have a function that is a quadratic trinomial with a positive lead coefficient, i.e. the equation of a convex up parabola.
If this is an algebra problem, use the formula for the -coordinate of the vertex of \ =\ ax^2\ +\ bx\ +\ c) , namely  to get your answer.
If this is a calculus problem, take the first derivative and set it equal to zero. Solve the resulting equation. Verify that the second derivative evaluates to a positive number at this value of the independent variable.
John
My calculator said it, I believe it, that settles it
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