Question

The famous Kate Lynn Horsefeed is building a box as part of her science project. It is to be built from a rectangular piece of cardboard measuring 25 cm by 40 cm by cutting out a square from each corner and then bending up the sides. Find the size of the corner square which will produce a container that will hold the most.

My question is how do you even begin to form the equation?

Answer 1

So you are doing related rates I see?

First imagine it:




The \(x\) variable is the length of the square she is cutting from each corner. The image shows the 'geometric net' of her box. The length that she cuts is the same as the height:




As you can see, it is asking for the value of x that will carry the greatest volume.
The volume of a box is

\(V = lwh\)

And in our scenario
\(l = 40-2x\)
\(w = 25-2x\)
\(h = x\)

Why did I subtract 2x from the l and w? Because if you look at our net, cutting the x out would lower the length or width on either side.

Now, substitute into the volume and we get a function of l, w, and h values to yield the exact possible volumes, correct?


\(V(x) = x(40-2x)(25-2x)\)

Simplify
\(V(x)=4x^3 - 130x^2 + 1000x\)

To find the maximum, you would take the derivative and set it to 0 to find the extrema. Then you would evaluate whether it is a minimum or maximum by plugging it into the second derivative. a negative value = maximum. positive = minimum

Answer 2

\[\frac{ d }{ dx }(V(x) = 4x^3 - 130x^2 + 1000x)\]
\(V'(x) = 12x^2 - 260x + 10000\)
Extrema at 50/3 and 5.

\(V''(x) = 24x - 260\)
\(V''(5)<0\)
\(V''(50/3)>0\)

Thus the value of x to yield the greatest volume is 5 cm

Answer 3

Thank you so much, dude! That really helps out a ton.
And this is Calculus AB semester long. Taking Calculus BC semester long after finishing AB, so we're being rushed into things and trying to catch up is pretty difficult I feel.

Answer 4

Np dude.