# Is this quadratic word problem correct so far?

I'm a little confused as to how to solve this word problem I have. The problem is: A rectangular box (with a top) has a square base. The sum of the lengths of its edges is 8 feet. What dimensions should the box have in order for its surface area to be as large as possible?

My work so far is:

x = Length of side of base (ft) y = Height of rectangular box (ft) M = Surface area of box (ft2) The question asks for the largest surface area possible with a rectangular box that has a square base and the sum of whose edges is 8 ft. To determine that, we must first determine the formulas with the given information for the sum of the edges and the surface area. The area of a square with side length x is x2, and the area of a rectangle with side lengths x and y is xy. 4y + 8x = 8 4y = 8 – 8x y = 2 – 2x M = 2x2 + 4xy M = 2x2 + 4x(2-2x) M = -6x2 + 8x
And that is the quadratic equation for the surface area

Also, for the bottom line, if that is truly the equation, would that be considered the equation for the surface area, or should I reword it to something else? Thanks!

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It is a correct expression for surface area in terms of $x$. Perhaps the range of validity should be mentioned. –  André Nicolas Mar 9 '13 at 18:32
What do you mean? –  Someone Mar 9 '13 at 18:59
I mean that for completeness you should specify that $0\lt x\lt 1$, or perhaps, if you allow "degenerate" boxes, $0\le x\le 1$. In a number of max/min problems, it is important to be aware of what the allowed values of $x$ are. –  André Nicolas Mar 9 '13 at 19:04
I'm confused too! What "lengths of its edges" have to add to 8 feet? The three dimensions of the box? The sides of the square base? The total of all 12 edges? –  DJohnM Mar 9 '13 at 21:27

You want to maximize the surface area, so

$$M(x) = -6x^2 + 8x \implies M'(x) = -12x + 8$$

Now, $$M'(x) = 0 \implies -12x + 8 = 0 \implies x = \frac{8}{12} = \frac{2}{3}$$

So your surface area is a maximum when $x = \frac{2}{3}$ and thus $y = 2\left(1-\frac{2}{3}\right) = \frac{2}{3}$

So essentially you want a cube.

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Great, thank you! I was just a little confused as to whether or not 2/3 was a correct value, but this is great! –  Someone Mar 9 '13 at 18:48
This could also be answered without calculus. The quadratic $6x^2-8x+M=0$ has real solution(s) for x only if the discriminant is non-negative. The biggest possible M leads to a zero discriminant and thus $x=\frac{8}{12}$ –  DJohnM Mar 9 '13 at 21:44