A box is to be constructed from a sheet of cardboard that is 20 cm by 60 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?

- zeesbrat3

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- zeesbrat3

@ganeshie8

- zeesbrat3

@jim_thompson5910

- zeesbrat3

@freckles

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- jim_thompson5910

|dw:1436828452402:dw|

- jim_thompson5910

if we make the cuts and fold up the box, we get something like this
|dw:1436828568511:dw|

- zeesbrat3

Okay. What next? I know Volume would be lwh --- so x(20-2x)(60-2x)?

- jim_thompson5910

correct

- zeesbrat3

So, then what?

- jim_thompson5910

ok so if you graph x(20-2x)(60-2x) with something like desmos, you get this (see attached)

##### 1 Attachment

- zeesbrat3

Okay?

- jim_thompson5910

##### 1 Attachment

- zeesbrat3

So 2525 is the max volume?

- jim_thompson5910

desmos is saying that the max point is (4.514, 2525)
so x = 4.514 and the max volume is 2525
yes

- zeesbrat3

Thank you!

- jim_thompson5910

no problem

- zeesbrat3

So what if you wanted to work it out by hand? Just wondering..

- jim_thompson5910

to do so by hand, you would need to know calculus (specifically, derivatives)

- zeesbrat3

I do haha. I did it originally using derivatives and my teacher yelled at me and marked it wrong

- jim_thompson5910

such a strange thing to be yelled at for lol

- jim_thompson5910

I'm guessing you're in precalculus and the teacher wanted you to use technology like a graphing calculator

- zeesbrat3

I'm actually in calculus and my sister taught me derivatives

- jim_thompson5910

ok let's use derivatives then

- jim_thompson5910

first expand out x(20-2x)(60-2x) to get
x(20-2x)(60-2x) = x(1200 - 160x + 4x^2) = 4x^3 - 160x^2 + 1200x

- jim_thompson5910

then you apply a derivative to 4x^3 - 160x^2 + 1200x to get
y = 4x^3 - 160x^2 + 1200x
dy/dx = 3*4x^(3-1) - 2*160x^(2-1) + 1*1200x^(1-1)
dy/dx = 12x^2 - 320x + 1200

- zeesbrat3

Ya, I got that far. Then quadratic?

- jim_thompson5910

Now you set the derivative 12x^2 - 320x + 1200 equal to zero and solve for x
12x^2 - 320x + 1200 = 0
x = 22.15250437 or x = 4.514162294
I skipped a bunch of steps since I'm sure you know the quadratic formula very well at this point

- zeesbrat3

Then plug them back in for the y

- jim_thompson5910

since the width is 20-2x, this means that 20-2x must be positive
20 - 2x > 0
20 > 2x
2x < 20
x < 10
so x is both positive, x > 0, and less than 10.
Put together, 0 < x < 10 is the domain restriction here. This is visually confirmed in that desmos graph above

- jim_thompson5910

because 0 < x < 10, we can ignore x = 22.15250437

- zeesbrat3

I find it extremely frightening that you just completed the problem so quickly going the long way

- jim_thompson5910

I've had lots of practice

- zeesbrat3

Maybe one day I will be able to do that

- jim_thompson5910

you will with enough practice

- zeesbrat3

Again, thank you for your help

- jim_thompson5910

no problem

Looking for something else?

Not the answer you are looking for? Search for more explanations.