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anonymous

  • 5 years ago

It is your job to design an open-topped rectangular box with height h and a square base of side length s using 5 sheets of metal (one for the base and 4 equal ones for the sides). You are allowed to use a total of 5 square meters of metal. Let V be the volume of the box and A be the area of the box. Begin by finnding formulas for V and A in terms of s and h. Given that A = 5 and a formula for V involving ONLY s (no h's). Finally use your graphing calculator to estimate the value of s that gives the maximum volume. s is

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  1. anonymous
    • 5 years ago
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    I'll write something up for you and scan it, okay?

  2. anonymous
    • 5 years ago
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  3. anonymous
    • 5 years ago
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    You have to plug in A=5 and plot the equation, \[V=\frac{s}{4}(5-s)\]

  4. anonymous
    • 5 years ago
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  5. anonymous
    • 5 years ago
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    You're being asked to find the highest point on a plot, when you punch in the formula into your calculator. The horizontal axis is s (side length) and the vertical axis is V, the volume that your box has for a given s. You should see that, as s increases from 0, so does the volume...up to a certain point...then the volume equation comes down again even though the side length is still increasing.

  6. anonymous
    • 5 years ago
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    You have to find the maximum point with your calculator.

  7. anonymous
    • 5 years ago
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    You should find that the maximum volume occurs for s=2.5.

  8. anonymous
    • 5 years ago
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    Okay?

  9. radar
    • 5 years ago
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    I am trying to follow the Volume.jpg. where you transition from:\[V=(s ^{2}(A-s ^{2}))/4s\] to\[As/4-s ^{2}/4\] I am coming up with \[As/4-s ^{3}/4\] Where am I going wrong?

  10. anonymous
    • 5 years ago
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    \[A=4 \times (side.area)+(area.of.base)= 4sh+s^2\]so\[h=\frac{A-s^2}{4s}\]Since\[V=s^2h \rightarrow V=s^2.\frac{A-s^2}{4s}=\frac{s}{4}(A-s^2)\]

  11. anonymous
    • 5 years ago
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    You weren't going wrong.

  12. radar
    • 5 years ago
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    O.K Just looks different LOL

  13. anonymous
    • 5 years ago
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    I dropped a superscript on paper.

  14. anonymous
    • 5 years ago
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    No, there's an error in what I derived.

  15. radar
    • 5 years ago
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    I like the use of attachments

  16. anonymous
    • 5 years ago
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    When they're correct... :(

  17. radar
    • 5 years ago
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    Thanks for clearing things up for me. When you are treading on new ground, sometimes you lose your confidence. I was unsure, thanks again.

  18. anonymous
    • 5 years ago
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    Dscribblez, ignore the scans...the equations typed above are correct. You need to plug the one for V into your calculator and estimate the s that gives you the greatest positive value of V.

  19. anonymous
    • 5 years ago
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    No worries, radar.

  20. anonymous
    • 5 years ago
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  21. anonymous
    • 5 years ago
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    The maximum value occurs at the intersection of the line and the cubic. The exact s-value is\[s=\sqrt{\frac{5}{3}}\]but I'm thinking if you quote that, your teacher might think you've cheated. Plot it, estimate it. Good luck.

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