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Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, $ 3 ft $ wide, by cutting out a square from each other of the four corners and bending up the sides. Find the largest volume that such a box can have.

(a) Draw several diagrams to illustrate the situation, some short boxes with large bases. Find the volumes of several such boxes. Does it appear that there is a maximum value? If so, estimate it.

(b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols.

(c) Write an expression for the volume.

(d) Use the given information to write an equation that relates the variables.

(e) Use part (d) to write the volume as a function of one variable.

(f) Finish solving the problem and compare the answer with your estimate in part (a).

a) max volume appears to be 2 $\mathrm{ft}^{3}$

b) $3-2 x$

c) $V=x(3-2 x)^{2}$

d) $y=3-2 x$

e) $V(x)=x(3-2 x)^{2}$

f) $x=\frac{1}{2}$ ft produces the maximum volume of 2 $\mathrm{ft}^{3}$

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Sharieleen A.

October 23, 2020

Thought I needed a tutor to help with Calculus: Early Transcendentals , but this helps a lot more.

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October 27, 2020

Thought I needed a tutor to help with Calculus: Early Transcendentals, but this helps a lot more.

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August 4, 2021

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So for this problem we have a box Um and it's 3ft wide. We cut out squares from the four corners. So we have um that are volume is equal to, Since we know the dimensions are square, it's going to be 3 -2 x. The volume is 3 -2 x. I'm 3 -2X. So we can just square that And then that can be times X. So this is going to end up giving us a graph and we can maximize the volume which we see is going to be right here at two ft, cute. So that would be maximized volume. And then we want to draw a diagram to illustrate the general situation what is pictured as being this square piece right here. And then we cut out some corners. We know the distance is X in the corners. So that means the total length is now going to be the original length minus two X. Because we cut out the two squares with length X. Um and ultimately were able to maximize this function um maximizing the area. We see that there are restrictions on this because it comes to a point to where you obviously can't have. Um 1.5 is the most the X could be because then you would no longer have any side length, so it has to be between zero and 1.5, but not including those values.

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