Question 199523
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I have to presume that you are a calculus student and are familiar with the process of finding the first and second derivatives of a polynomial function because that is the only way I know to solve this problem.


Another assumption that I have to make is that by "largest rectangular box" you mean the "rectangular box with the largest volume".


Given all of that:


The restriction is that the sum of the dimensions be not exceed 3 m and that at least one pair of opposite faces of the rectangular box are squares.


Let *[tex \Large x] represent the measure of the side of one of the square faces.  Then we can say that the remaining dimension is *[tex \Large 3 - 2x].


The volume is the product of the three dimensions so the volume function with respect to the dimension of the square end is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(x) = x^2(3 - 2x) = -2x^3 + 3x^2]


The feasible domain for *[tex \Large x] in this situation is *[tex \Large (0,1.5)] because if *[tex \Large x = 0] then you obviously have a zero volume box and if *[tex \Large x = 1.5] then the length must be zero because 2 times 1.5 is 3, and again you have a zero volume box.


We are interested in finding a local maximum of the volume function on the interval (0,1.5).


Take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V'(x) = -6x^2 + 6x]


Set the first derivative equal to zero and solve (Fermat's Theorem)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -6x^2 + 6x = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -6x(x - 1) = 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 0]


which must be excluded because it is not in the feasible interval, or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 1]


Now that we know that there is a critical point at *[tex \Large x = 1] we need to apply the second derivative test to determine if it is a maximum, minimum, or a possible inflection point.


Take the second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V''(x) = -12x + 6]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V''(1) = -12(1) + 6 < 0]


Therefore V(1) is a local maximum.  That means that the measure of each side of the square end of the box must be 1, and since 3 - 2 = 1, the measure of the length must be 1 as well.  The maximum volume rectangular box for a given sum of dimensions is a cube.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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