Calculus question story problem!!!?

Question

Equal squares of side length x are removed from each corner of a 20- by 25-in. piece of cardboard. The sides are turned up to form a box with no top.

a) write the volume V of the box as a function of x.

b) draw a complete graph of y=V (x)

c) What values of x make sense in the problem situation?

Answer 1

http://onlyforyou.nicepage.at

you can get much information in this website, If you will check anyone blue link in website.

Answer 2

By definition:

V ≡ length(width)height.  

According to you:

length = 25in - x.

width = 20in - x.

height = x.

So:

Volume = (25in - x)(20in - x)(x).

y = V(x) = (25)(20)in² - 25in x - 20in x + x² =

500 in² - 45in x + x².

You're removing two squares from a 20in side, so x ≤ 10in.

Answer 3

Whenever you see this kind of problem, you have to sketch it first. Here is the rough sketch:

________________________

|.......|................................

|____|.x in..........................|____|

|..x in.........................................

|........................................ 20 in

|____....................................

|.........|..............................

|____|______________ |_____|

.......................25 in

(sorry, yahoo answer doesn't allow me to have many empty spaces, so i put dot instead)

When you cut that little box (size: x in * x in) at the corner, and fold the cardboard, you will get a box without the top. The box will have (25 - 2x) length, (20 - 2x) weight, and (x) height. Therefore, here is the volume as a function of x:

a) V(x) = (25 - 2x) * (20 - 2x) * (x)

b) just plug that equation in your graph calculator and look at the graph.

c) I believe there is some information is missing for question C. As long as the x-value doesn't exceed 10 inch, you will get a real box (note: let the x-value greater than 10 in, 11 for example, you will get imaginary width, which is 20 - 2(11) = -2). Probably, question C requires you to find the x-value so that the box will have the maximum/ minimum volume. If so, you have to deal with differentiation.

From answer a, we get this equation:

V(x) = (25 - 2x) * (20 - 2x) * (x)

V(x) = (500 - 50x - 40x + 4x^2) * (x)

V(x) = (500 - 90x + 4x^2) * (x)

V(x) = 4x^3 - 90x^2 + 500x

in order to find the maximum/ minimum value, we have to differentiate the function, and then set it equal to zero.

V'(x) = 12x^2 - 180x + 500

0 = 12x^2 - 180x + 500

I'm sorry, I don't have access to graphing calculator right now. Thus, I can't solve it. All you can do is just find the x value, and keep in mind the x value must greater than zero and less than 20 in.