A box is to be constructed from a sheet of cardboard that is 10 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides.
What is the maximum volume this box could have?

Question

A box is to be constructed from a sheet of cardboard that is 10 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides.
What is the maximum volume this box could have?

chris215 · Answer

I got 2.36

agent0smith · Answer

2.36 what?? cm? cm^3? 

The height of the box will be x. The length 50 - 2x. The width 10 - 2x.

So the volume, V = x(50 - 2x)(10 - 2x).

If you differentiated that and found the x value that gives maximum volume, then found the value of V using that x-value, you might be right.

But from looking at a graph of the volume equation, you aren't finished yet.

chris215 · Answer

2.36 cm ^3

agent0smith · Answer

I think x = 2.36 cm, that's all you've found. That isn't what the question asked for, though.

agent0smith · Answer

x is a length, not a unit of volume.

chris215 · Answer

v = x(10 - 2x)(50 - 2x) 
v = x(500 - 20x - 100x + 4x^2) 
v = 4x^3 - 120x^2 + 500x 
dv/dx = 12x^2 - 240x + 500 
x = (-(-240) +/- sqrt((-240)^2 - 4*12*500)) / (2*12) 
x = (240 +/- sqrt(57600 - 24000)) / 24 
x = (240 +/- sqrt(33600)) / 24 
x = 10 +/- sqrt(175/3) 
x = 10 - sqrt(175/3) = 2.36

agent0smith · Answer

Right. That's x. But again... the question did not ask for x.

chris215 · Answer

ok idk what to do ..

agent0smith · Answer

Tell me what the question asked you to find.

chris215 · Answer

find the maximum volume?

agent0smith · Answer

Yes. 

You have a volume equation. You now have the x-value that gives you the maximum volume.

mathstudent55 · Answer

|dw:1454480595043:dw|

$V = (50 - 2x)(10 - 2x)x$

$V = 4x(x - 25)(x - 5)$

$V = 4x(x^2 - 30x + 125$

$V = 4x^3 - 120x^2 + 500x$

$\dfrac {dV}{dx} = 12x^2 - 240x + 500$

$12x^2 - 240x + 500 = 0$

$3x^2 - 60x + 125 = 0$

$x = \dfrac{-(-60) \pm \sqrt{(-60)^2 - 4(3)(125)}}{2(3)} $

$x = \dfrac{60 \pm \sqrt{3600 - 1500}}{6} $

$x = \dfrac{60 \pm \sqrt{2100}}{6} $

$x = \dfrac{60 \pm 10\sqrt{21}}{6} $

$x = 10 \pm \dfrac{5\sqrt{21}}{3} \approx 17,64, 2.36$

We discard the larger number since it is larger than 10 cm, the width of the rectangular cardboard.

$x = 10 - \dfrac{5\sqrt{21}}{3} $ yields the largest volume.

Now insert this value of x in the volume formula and evaluate it.

$V = 4x^3 - 120x^2 + 500x$

$V = 4 \left( 10 - \dfrac{5\sqrt{21}}{3} \right) ^3 - 120 \left( 10 - \dfrac{5\sqrt{21}}{3} \right)^2 + 500\left( 10 - \dfrac{5\sqrt{21}}{3} \right)$

Now you can finish it.
Of course, if all you need is an approximate volume, then use x = 2.36