The length of a rectangular box is 3cm more than the width....?

Question

Can't remember how to do this, can someone explain and show work?

The length of a rectangular box is 3cm more than the width and the height is 1 cm more than half the width. The volume is 216 cm^3. Recall that V=lwh

a) Algebraically determine the dimensions of the box.

b) Graphically determine the dimensions of the box.

Answer 1

w = x

l = x+3

h = 1+0.5x

So we take the above values and plug into the equation

V = x(x+3)(0.5x+1)

216 = x(x+3)(0.5x+1)

try distributing...

216 = x(0.5x^2 + 2.5x + 3)

0 = 0.5x^3 + 2.5x^2 + 3x -216

The best we can do is try and factor the equation

0 = x^3 + 5x^2 + 6x -432

Try factoring the above if possible...

You got a doozie here, but you can supposedly factor the above using different possible factors...

I'll cheat right here...

Using the table feature on my graphing calculator...we see that x=6 in this equation, because we have to remember, a size can only be positive. So we want an only positive root. So lets do this...take the origional equation and long divide it by x-6 (the factor which will give us a solution of 6)

And x=6 is the only real root of the equation (the quadratic we are left with has two imaginary roots)

So now we end up with x = 6

So w=6

l = 9

h = 4

9*6*4 = 216

To do this graphically, just graph the third degree polynomial and look at where it crosses the x-axis, which would be x=6

Answer 2

If possible, choose the smallest dimension to be represented by an algebraic variable. In this case, let w = the width. The length would then be (w+3), and the height would be (w/2)+1. Now use the formula for the volume by substituting these values for the l, w, and h, and make the product equal to 216.

(w+3) * w * [(w/2)+1] = 216. Simplify the left side, and get
(w^2 + 3w) [(w/2)+1] = 216, then
w^3/2 + w^2 +3w^2/2 + 3w = 216.

Combine the middle terms, move the 216, and get
w^3/2 + 5w^2/2 + 3w - 216 = 0.
Multiply through by 2 to get rid of the denominators,
w^3 + 5w^2 + 6w - 432 = 0

That's as far as I can take it for you. You need the formula for solving third-degree equations at this point.

Answer 3

The question does not say to solve the question.
Usually, the question would say to solve for the maximum width or length or height. This question does not.

a)
w=width
l=length
h=height

w=w
l=w+3
h=w/2 +1


b) The answer is 6. How do you graphically show this?

Use the formula (1/2)w^3 + (5/2)w^2+3w=216 as explained in other answers.

Show volume along the side and w along the bottom.
w will go from minimum of 0 to maximum of 10

Volume=(1/2)w^3 + (5/2)w^2+3w

Calculate a maximum volume before you draw the graph.

I pick w=10 because 1) volume will be larger than 216 and 2) 10 is easy to calculate.

so maximum volume is 780.

Now draw the graph, w along the bottom going from 0 to 10 and Volume along the side going from 0 to 800.

Draw a horzontal line where V=216

Plot (w, V) (0,0) - (10,780)
calculate (5,V1) and then (6,216)
connect the plotted points
Ohh that looks so beautiful.
RB

Answer 4

let 'l' be the length, w the width & h the height of the box . volume is lwh
given :-
l=3+w
h=1+(w/2)
lwh=216
(3+w)w(1+w//2)=216
(3+w)(2+w)w=432
6+w^2+5w)w=432
w^3+5w^2+6w-432=0
solve this cubic eqn using synthetic method and u will get the ans.

Answer 5

l = w + 3
h = w/2 + 1
v = lwh
216 = (w + 3)(w)(w/2 + 1)
216 = (w^2 + 3w)(w/2 + 1)= W=w^3/2 +(1 + 3/2)w^2 + 3w
w^3/2 + 5w^2/2 + 3w = 216
This leads us to cubic equation which is hard to solve algebraically.

Answer 6

l = w +3
h = 1/2w+1
216 = (w+3)(1/2w+1)(w)

1/2w^3 + 5/2w^2 + 3w - 216 = 0

now factor and you get w = 6

w = 6
l = 9
h = 4

I'm going to bed now its 3 am and I had to do this w/o a calculator, I hope it helped. *YAWN*

Answer 7

let width xcm
then length =(x+3)cm
height =(x/2+1)cm
volume=216cm^3=l*W*h
=>x*(x+1)*(x+2)/2=216
solve for x and substitute in l,w,h