Expert: Paul Klarreich - 12/22/2009

Question
Hi

The question goes:

A rectangular box open at the top has a square base of side x metres. The total surface area of the box is 16m squared and the volume is Vm cubed. Show that V = 1/4 x(16-x squared)
What is the domain and range of the function V(x) in this context?

I understand the proof but how to work out the domain and range has me confused. I understand how to find the range of x but not how to work out the maximum value of V(x). I don't understand what d is in dV/dx = 1/4 (16-3x squared) = 0.

Thank you !

Answer
Questioner: Logice
Country: Australia
Category: Advanced Math
Private: No
Subject: d in functions
Question: Hi

The question goes:

A rectangular box open at the top has a square base of side x metres. The total surface area of the box is 16 m squared and the volume is Vm cubed.

Show that V = 1/4 x(16-x squared)
What is the domain and range of the function V(x) in this context?

I understand the proof but how to work out the domain and range has me confused. I understand how to find the range of x but not how to work out the maximum value of V(x). I don't understand what d is in dV/dx = 1/4 (16-3x squared) = 0.
........................................................

1. Show that V = 1/4 x(16-x squared)

That's not so hard:

S = 16 = x^2 + 4hx   << one base plus four sides.

h = (16 - x^2)/4x    << solve for h.

V = x^2 h = x^2(16 - x^2)/4x  << substitute your h.

V = x^2 h = x(16 - x^2)/4    << simplify a bit.

OK.

2. Domain of V(x) as a function is all x, but in this context, you don't want negative V's.  So:

V = x(4 - x)(4 + x)/4

So you cannot have negative x, and you cannot have  negative '4-x' either.

You must have  x >= 0, and  4-x >= 0, or x <= 4

That is it:  0 <= x <= 4

........................

3. ... how to work out the maximum value of V(x). I don't understand what d is in dV/dx = 1/4 (16-3x squared) = 0.

Sorry, mate -- for this you must wait until you start calculus.  See you then.