# Volume problem

• Feb 2nd 2009, 03:41 AM
requal
Volume problem
A rectangular sheet of metal measures 6cm by 8cm. Four equal squares are cut from each corner of the sheet (so it looks like a net of a rectangular prism, without the lid) and the remaining metal is bent to form an open ended rectangular tray. (so the equation of the volume would be V=4x(4-x)(3-x)]
I) What would be the largest possible value of x?
II) What would be the smallest possible value of x?

Is there a way to do this without calculus?
• Feb 2nd 2009, 05:07 AM
earboth
Quote:

Originally Posted by requal
A rectangular sheet of metal measures 6cm by 8cm. Four equal squares are cut from each corner of the sheet (so it looks like a net of a rectangular prism, without the lid) and the remaining metal is bent to form an open ended rectangular tray. (so the equation of the volume would be V=4x(4-x)(3-x)]
I) What would be the largest possible value of x?
II) What would be the smallest possible value of x?

Is there a way to do this without calculus?

The volume is calculated by:

\$\displaystyle V(x)=x(8-2x)(6-2x)\$

The three factors correspond to the edges of the box and must be positive (or zero).

Therefore the smallest possible length of x is x = 0, but then the volume is zero too. Actually you leave the metal sheet untouched.

The largest possible length of x is x = 3, but then the width of the box is zero and consequently the volume is zero too. Actually you fold the sheet of metal once.

Thus the domain of V is \$\displaystyle d_V=[0, 3]\$