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Holiday Gift Box

Holiday Gift Box. By Anthony, Samson, and Mitchell. Introducing. Samson Aghedo. Anthony Ambrosino. I’m counting on doing well in this class. To the making of books: there is no end. Too much studying is worrisome to the flesh. This is life so go and have a ball.

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Holiday Gift Box

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  1. Holiday Gift Box By Anthony, Samson, and Mitchell

  2. Introducing Samson Aghedo Anthony Ambrosino I’m counting on doing well in this class.. To the making of books: there is no end. Too much studying is worrisome to the flesh. This is life so go and have a ball. Because the world don't move to the beat of just one drum. What might be right for you may not be right for some. You take the good, you take the bad, you take them both and there you have ... my opening statement. Mitchell Taylor

  3. The Problem A charity organization is given ten sheets of square cardboard, length 4ft, for boxing up holiday gifts for children. The organization wants to create a open-topped box with the largest volume by cutting four equal sized square pieces from the four corners of the cardboard sheets and folding up the sides. Help the charity group by finding the largest possible volume and its corresponding side lengths.

  4. Possible Layouts

  5. The Diagram 4ft 4ft

  6. The Approach • Formulate an equation for the volume to be optimized. In this case, the volume is determined to be V=xy² with respect to the diagram. • We can also see that since the sides are 4ft long, 2x+y=4ft. • Solve the second equation for y and plug in result to the first equation.

  7. The Work y=4-2x V=x(4-2x)²=4x³-16x²+16x V’=12x²-32x+16 0=12x²-32x+16 0=3x²-8x+4 x=2, 2/3 • Substitute the new y-value into the volume equation and expand. • Differentiate the new • volume equation and • Solve for x.

  8. The Work (Continued) y=4-2x=4-2(2)=0ft V=(1)(0)²=0ft² y=4-2(2/3)=8/3ft V=(2/3)(8/3)²=4.74ft³ Largest Possible Volume: 4.74ft³ when x=2/3ft, y=8/3ft Now use the generated x-values to solve for their corresponding y-values. Then plug them both into the volume equation and solve. After plugging in resulting x and y- values we see that the largest volume of 4ft² is achieved when x=1ft and y=2ft.

  9. END

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