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anonymous
 5 years ago
It is your job to design an opentopped 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
anonymous
 5 years ago
It is your job to design an opentopped 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll write something up for you and scan it, okay?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to plug in A=5 and plot the equation, \[V=\frac{s}{4}(5s)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to find the maximum point with your calculator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should find that the maximum volume occurs for s=2.5.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I am trying to follow the Volume.jpg. where you transition from:\[V=(s ^{2}(As ^{2}))/4s\] to\[As/4s ^{2}/4\] I am coming up with \[As/4s ^{3}/4\] Where am I going wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[A=4 \times (side.area)+(area.of.base)= 4sh+s^2\]so\[h=\frac{As^2}{4s}\]Since\[V=s^2h \rightarrow V=s^2.\frac{As^2}{4s}=\frac{s}{4}(As^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You weren't going wrong.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0O.K Just looks different LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dropped a superscript on paper.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, there's an error in what I derived.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I like the use of attachments

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When they're correct... :(

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for clearing things up for me. When you are treading on new ground, sometimes you lose your confidence. I was unsure, thanks again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Dscribblez, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The maximum value occurs at the intersection of the line and the cubic. The exact svalue 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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