Question 201030
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There is no correct area on this site for this problem because it is a Calculus local maximum problem.  The closest thing I could find was <i>Algebra: Polynomials, rational expressions, and equations</i>


Let *[tex \Large x] represent the measure of the side of the squares cut out of the corners.  That means that the width and length of the resulting box must measure *[tex \Large 30 - 2x] and the height of the resulting box is just *[tex \Large x].  So you can write a function representing the volume of the box in terms of the dimension of the cutout square thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(x) = x(30 - 2x)(30 - 2x) = 4x^3 - 120x^2 + 900x]


Given a continuous and differentiable function, such as the above polynomial function, local extrema can be found at the value of the independent variable for which the value of the first derivative is equal to zero.  So, take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dx} = 12x^2 - 240x + 900]


Set the first derivative equal to zero and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x^2 - 240x + 900 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3x - 45)(4x - 20) = 0]


Hence, the *[tex \Large x]-coordinates of the local extrema are 5 and 15.


Given a continuous and twice-differentiable function if the 2nd derivative is less than zero evaluated at the *[tex \Large x]-coordinate of an extrema, that extrema is a maximum.  If it is positive, the extrema is a minimum.


Take the 2nd derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2V}{dx^2} = 24x - 240]


Then evaluate at 5:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V''(5) = 24(5) - 240 = -120]


So 5 is the *[tex \Large x]-coordinate of a maximum.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V''(15) = 24(15) - 240 = 120]


So 15 is the *[tex \Large x]-coordinate of a minimum.  Which, by the way, makes very good sense because if you cut 15 inch squares out of the corners of a 30 by 30 piece of material, you would end up with a 15 X 0 X 0 box having zero volume.


So, you cut 5 inch squares from each corner, leaving you with a 5 X 20 X 20 box.


Here's a picture of your volume function:


{{{drawing(
500, 500, -1, 17, -100, 2100,
grid(1),
green(circle(5,2000,.2)),
graph(
500, 500, -1, 17, -100, 2100,
4x^3-120x^2+900x
))}}}



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