zeesbrat3
  • zeesbrat3
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zeesbrat3
  • zeesbrat3
@ganeshie8
zeesbrat3
  • zeesbrat3
@jim_thompson5910
zeesbrat3
  • zeesbrat3
@freckles

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More answers

jim_thompson5910
  • jim_thompson5910
|dw:1436828452402:dw|
jim_thompson5910
  • jim_thompson5910
if we make the cuts and fold up the box, we get something like this |dw:1436828568511:dw|
zeesbrat3
  • zeesbrat3
Okay. What next? I know Volume would be lwh --- so x(20-2x)(60-2x)?
jim_thompson5910
  • jim_thompson5910
correct
zeesbrat3
  • zeesbrat3
So, then what?
jim_thompson5910
  • jim_thompson5910
ok so if you graph x(20-2x)(60-2x) with something like desmos, you get this (see attached)
zeesbrat3
  • zeesbrat3
Okay?
jim_thompson5910
  • jim_thompson5910
zeesbrat3
  • zeesbrat3
So 2525 is the max volume?
jim_thompson5910
  • 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
  • zeesbrat3
Thank you!
jim_thompson5910
  • jim_thompson5910
no problem
zeesbrat3
  • zeesbrat3
So what if you wanted to work it out by hand? Just wondering..
jim_thompson5910
  • jim_thompson5910
to do so by hand, you would need to know calculus (specifically, derivatives)
zeesbrat3
  • zeesbrat3
I do haha. I did it originally using derivatives and my teacher yelled at me and marked it wrong
jim_thompson5910
  • jim_thompson5910
such a strange thing to be yelled at for lol
jim_thompson5910
  • jim_thompson5910
I'm guessing you're in precalculus and the teacher wanted you to use technology like a graphing calculator
zeesbrat3
  • zeesbrat3
I'm actually in calculus and my sister taught me derivatives
jim_thompson5910
  • jim_thompson5910
ok let's use derivatives then
jim_thompson5910
  • 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
  • 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
  • zeesbrat3
Ya, I got that far. Then quadratic?
jim_thompson5910
  • 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
  • zeesbrat3
Then plug them back in for the y
jim_thompson5910
  • 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
  • jim_thompson5910
because 0 < x < 10, we can ignore x = 22.15250437
zeesbrat3
  • zeesbrat3
I find it extremely frightening that you just completed the problem so quickly going the long way
jim_thompson5910
  • jim_thompson5910
I've had lots of practice
zeesbrat3
  • zeesbrat3
Maybe one day I will be able to do that
jim_thompson5910
  • jim_thompson5910
you will with enough practice
zeesbrat3
  • zeesbrat3
Again, thank you for your help
jim_thompson5910
  • jim_thompson5910
no problem

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