Question 549580
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Since *[tex \Large x] is the perimeter of the first rectangle where the length is twice the width, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6w\small{_1}\LARGE\ =\ x]


From which we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\small{_1}\LARGE\ =\ \frac{x}{6}]


Likewise, since the perimeter of the other rectangle is *[tex \Large 16\ -\ x] and the length of the second rectangle is 3 times the width, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8w\small{_2}\LARGE\ =\ 16\ -\ x]


From which we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\small{_2}\LARGE\ =\ \frac{16\ -\ x}{8} ]


The area of the first rectangle in terms of its width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w\small{_1}\LARGE\ \cdot\ w\small{_1}\LARGE\ =\ 2w\small{_1}\LARGE^2]


And the area of the second rectangle in terms of its width is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3w\small{_2}\LARGE\ \cdot\ w\small{_2}\LARGE\ =\ 3w\small{_2}\LARGE^2]


Then the total area is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w\small{_1}\LARGE^2\ +\ 3w\small{_2}\LARGE\,^2]


Substituting the earlier derived expressions for *[tex \Large w\small{_1}\LARGE] and *[tex \Large w\small{_2}\LARGE]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ 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 *[tex \Large x]-coordinate of the vertex of *[tex \Large f(x)\ =\ ax^2\ +\ bx\ +\ c], namely *[tex \Large x_v\ =\ \frac{-b}{2a}] 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
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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