Question 421518: Squares with sides of length x are cut from the corners of a rectangular piece of sheet metal with dimensions of 6 in by 10 in. The metal is then folded to make a open top box. What is the maximum volume of the box and what are its dimensions.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the length of the box will be 10 - 2x
the width of the bot will be 6 - 2x
the height of the box will be x
the volume of the box will be (10 - 2x) * (6 - 2x) * x
the value of x has to be less than 3 since if it was equal to 3, then the width of the box would be equal to 0.
the value of x can be 0, so we have a domain where 0 <= x < 3.
if we multiply all factors together, we get:
y = 4x^3 - 32x^2 + 60x
if we graph this equation, we get:
within our domain space, it looks like we will get a maximum value of y at somewhere around y = 30 when x is somewhere around x = 1.
to solve for the roots of this equation, we set it equal to 0.
we get 4x^3 - 32x^2 + 60x = 0
we factor out an x and a 4 to get:
4 * x * (x^2 - 8x + 15) = 0
this equation will have roots at x = 0, x = 3, x = 5
the root at x = 5 is outside our domain of 0 <= x < 3, so we are left with roots at x = 0 and x = 3
to find the max / min point of this cubic equation, we have to resort to calculus (i don't know any other way).
the equation is y = 4x^3 - 32x^2 + 60x
the derivative of the equation with respect to x is:
dy/dx = 12x^2 - 64x + 60
this derivative is the slope of the equation at a particular value of x.
we want the slope of the equation to be equal to 0 which would indicate a max / min point of the original equation.
we set dy/dx equal to 0 to get:
12x^2 - 64x + 60 = 0
since this is a standard quadratic equation in the form of ax^2 + bx + 2 = 0, we can solve for the roots of this equation by setting:
a = 12
b = -64
c = 60
from that, we derive the piece parts of the quadratic formula of:
2a = 24
-b = 64
b^2 = 4096
4ac = 2880
b^2 - 4ac = 2880
we then plug these values into the quadratic formula of x = (-b +/- sqrt(b^2-4ac))/2a to get:
x = (64 +/- sqrt(2880))/24
this gets us x = 4.119632981 or x = 1.213700352
x = 4.... is outside our domain so we are left with x = 1.213700352.
when x = 1.213700352, the equation of y = 4x^3 - 32x^2 + 60x becomes y = 32.83528294.
if we add a value of y = 32.83528294 to our equation that we previously graphed, we should see that it is at the maximum value of the equation, as shown below:
the solution to your problem is that the maximum volume of the rectangular solid is 32.83528294 cubic inches.
this occurs when the value of x is 1.213700352 inches.
i'm far from a calculus guru, but i knew enough to solve this problem using it.
hopefully this is in a class where you use calculus.
i didn't know how to solve for the maximum volume any other way.
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