Optimization Problem - # 2

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Jason76

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If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation x2+4xy=\displaystyle x^{2} + 4xy = Surface area, then use Volume equation l∗w∗h\displaystyle l * w * h or x2y\displaystyle x^{2}y and then back to surface area equation after doing the derivative on the volume equation)

x2+4xy=\displaystyle x^{2} + 4xy = Surface area

x2+4xy=10,800\displaystyle x^{2} + 4xy = 10,800

x24x+4xy4x=10,8004x\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}

x24x+y=10,8004x\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}

y=10,8004x−x24x\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x} :confused: On the right track here?
 
Yes. Is there a reason to keep writing "x24x\displaystyle \dfrac{x^2}{4x}" rather than just x4\displaystyle \dfrac{x}{4}?
 
If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation x2+4xy=\displaystyle x^{2} + 4xy = Surface area, then use Volume equation l∗w∗h\displaystyle l * w * h or x2y\displaystyle x^{2}y and then back to surface area equation after doing the derivative on the volume equation)

x2+4xy=\displaystyle x^{2} + 4xy = Surface area

x2+4xy=10,800\displaystyle x^{2} + 4xy = 10,800

x24x+4xy4x=10,8004x\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}

x24x+y=10,8004x\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}

y=10,8004x−x24x\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x}

y=10,800−x24x\displaystyle y = \dfrac{10,800 - x^{2}}{4x} - continuing on...
x2(10,800−x24x)\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x})

V=x2y\displaystyle V = x^{2}y

V=10,800x2−x44x\displaystyle V = \dfrac{10,800x^{2} - x^{4}}{4x}
 
Last edited:
Jason76 said:
If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation x2+4xy=\displaystyle x^{2} + 4xy = Surface area, then use Volume equation l∗w∗h\displaystyle l * w * h or x2y\displaystyle x^{2}y and then back to surface area equation after doing the derivative on the volume equation)

x2+4xy=\displaystyle x^{2} + 4xy = Surface area

x2+4xy=10,800\displaystyle x^{2} + 4xy = 10,800

x24x+4xy4x=10,8004x\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}

x24x+y=10,8004x\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}

y=10,8004x−x24x\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x}

y=10,800−x24x\displaystyle y= \dfrac{10,800 - x^{2}}{4x}- continuing on...

x2(10,800−x24x)\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x}) ← Where did that come from?? why did it come?
Click to expand...

You should have written (instead of depending on our mind-reading capabilities):

V(olume) = x2 ∗ 10,800−x24x\displaystyle V(olume) \ = \ x^{2} \ * \ \dfrac{10,800 - x^{2}}{4x}

Now what........
 
Last edited by a moderator:
If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation x2+4xy=\displaystyle x^{2} + 4xy = Surface area, then use Volume equation l∗w∗h\displaystyle l * w * h or x2y\displaystyle x^{2}y and then back to surface area equation after doing the derivative on the volume equation)

x2+4xy=\displaystyle x^{2} + 4xy = Surface area

x2+4xy=10,800\displaystyle x^{2} + 4xy = 10,800

x24x+4xy4x=10,8004x\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}

x24x+y=10,8004x\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}

y=10,8004x−x24x\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x}

y=10,800−x24x\displaystyle y = \dfrac{10,800 - x^{2}}{4x} - continuing on...

V=x2y\displaystyle V = x^{2}y - plug y into volume equation

x2(10,800−x24x)\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x})

V=10,800x2−x44x\displaystyle V = \dfrac{10,800x^{2} - x^{4}}{4x}

V′=4x(21600x−4x3)−(10,800x2−x4)(4)(4x)2\displaystyle V' = \dfrac{4x(21600x - 4x^{3}) - (10,800x^{2} - x^{4})(4)}{(4x)^{2}}

V′=864,000x2−16x4−43200x2+4x44x2\displaystyle V' = \dfrac{864,000 x^{2} - 16x^{4} - 43200x^{2} + 4x^{4}}{4x^{2}}

V′=820,800x2+20x44x2\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}}

V′=820,800x2+20x44x2=0\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}} = 0

V′=820,800x2+20x4=0\displaystyle V' = 820,800 x^{2} + 20x^{4} = 0

V′=820,800x2=−20x4\displaystyle V' = 820,800 x^{2} = -20x^{4}

V′=41,000x−2\displaystyle V' = 41,000 x^{-2}

On the right track here?



 
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Jason76 said:
If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation x2+4xy=\displaystyle x^{2} + 4xy = Surface area, then use Volume equation l∗w∗h\displaystyle l * w * h or x2y\displaystyle x^{2}y and then back to surface area equation after doing the derivative on the volume equation)

x2+4xy=\displaystyle x^{2} + 4xy = Surface area

x2+4xy=10,800\displaystyle x^{2} + 4xy = 10,800

x24x+4xy4x=10,8004x\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}

x24x+y=10,8004x\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}

y=10,8004x−x24x\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x}

y=10,800−x24x\displaystyle y = \dfrac{10,800 - x^{2}}{4x} - continuing on...

V=x2y\displaystyle V = x^{2}y - plug y into volume equation

x2(10,800−x24x)\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x})

V=10,800x2−x44x\displaystyle V = \dfrac{10,800x^{2} - x^{4}}{4x}

V′=4x(21600x−4x3)−(10,800x2−x4)(4)(4x)2\displaystyle V' = \dfrac{4x(21600x - 4x^{3}) - (10,800x^{2} - x^{4})(4)}{(4x)^{2}}

V′=864,000x2−16x4−43200x2+4x44x2\displaystyle V' = \dfrac{864,000 x^{2} - 16x^{4} - 43200x^{2} + 4x^{4}}{4x^{2}}

V′=820,800x2+20x44x2\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}}

V′=820,800x2+20x44x2=0\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}} = 0

V′=820,800x2+20x4=0\displaystyle V' = 820,800 x^{2} + 20x^{4} = 0

V′=820,800x2=−20x4\displaystyle V' = 820,800 x^{2} = -20x^{4}

V′=41,000x−2\displaystyle V' = 41,000 x^{-2}

On the right track here?.......................NO!
Click to expand...

V′=820,800x2+20x44x2=0\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}} = 0

820,800+20x2=0\displaystyle 820,800 + 20x^{2} = 0 ...............and continue....
 
Much more direct way....

V(olume) = x2 ∗ 10,800−x24x=2700∗x−x34\displaystyle V(olume) \ = \ x^{2} \ * \ \dfrac{10,800 - x^{2}}{4x} = 2700*x - \dfrac{x^3}{4}

V′=2700−3∗x24=0\displaystyle V' = 2700 - \dfrac{3*x^2}{4} = 0

2700=3∗x24\displaystyle 2700 = \dfrac{3*x^2}{4}

x=60\displaystyle x = 60

 
Subhotosh Khan said:
Much more direct way....

V(olume) = x2 ∗ 10,800−x24x=2700∗x−x34\displaystyle V(olume) \ = \ x^{2} \ * \ \dfrac{10,800 - x^{2}}{4x} = 2700*x - \dfrac{x^3}{4}

V′=2700−3∗x24=0\displaystyle V' = 2700 - \dfrac{3*x^2}{4} = 0

2700=3∗x24\displaystyle 2700 = \dfrac{3*x^2}{4}

x=60\displaystyle x = 60
Click to expand...

Hello, bit late but this gives a different answer to your previous comment. Which one is the correct way?

Also if you go down from this: "820,800+20x2=0820,800+20x2=0 ...............and continue.... " It does not give you 60... so not sure what to do here?
 
magdasykes said:
Hello, bit late but this gives a different answer to your previous comment. Which one is the correct way?

Also if you go down from this: "820,800+20x2=0820,800+20x2=0 ...............and continue.... " It does not give you 60... so not sure what to do here?
Click to expand...
That was a continuation of Jason's incorrect work. Discard that!

Follow the development from response 7.
 
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