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zeesbrat3
 one year ago
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
 one year ago
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?

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436828452402:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1if we make the cuts and fold up the box, we get something like this dw:1436828568511:dw

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Okay. What next? I know Volume would be lwh  so x(202x)(602x)?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1ok so if you graph x(202x)(602x) with something like desmos, you get this (see attached)

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So 2525 is the max volume?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1desmos is saying that the max point is (4.514, 2525) so x = 4.514 and the max volume is 2525 yes

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So what if you wanted to work it out by hand? Just wondering..

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1to do so by hand, you would need to know calculus (specifically, derivatives)

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1I do haha. I did it originally using derivatives and my teacher yelled at me and marked it wrong

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1such a strange thing to be yelled at for lol

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm guessing you're in precalculus and the teacher wanted you to use technology like a graphing calculator

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1I'm actually in calculus and my sister taught me derivatives

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1ok let's use derivatives then

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1first expand out x(202x)(602x) to get x(202x)(602x) = x(1200  160x + 4x^2) = 4x^3  160x^2 + 1200x

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then you apply a derivative to 4x^3  160x^2 + 1200x to get y = 4x^3  160x^2 + 1200x dy/dx = 3*4x^(31)  2*160x^(21) + 1*1200x^(11) dy/dx = 12x^2  320x + 1200

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Ya, I got that far. Then quadratic?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Now 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
 one year ago
Best ResponseYou've already chosen the best response.1Then plug them back in for the y

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1since the width is 202x, this means that 202x 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
 one year ago
Best ResponseYou've already chosen the best response.1because 0 < x < 10, we can ignore x = 22.15250437

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1I find it extremely frightening that you just completed the problem so quickly going the long way

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I've had lots of practice

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Maybe one day I will be able to do that

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you will with enough practice

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Again, thank you for your help
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