Consider rectangles located in the first quadrant and inscribed under an increasing curve, with the upper right hand corner vertical line x = 3 and the upper left hand corner on the curve y = x^(1/3).

Find the position, width, height and area of the largest such rectangle Please. Thanks

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    $\begingroup$ You've asked several questions without detailing your effort. I'm very supportive of having an online community to help clarify problems you're having with an exercise, but this is not a personal homework service. $\endgroup$ – Clayton Mar 27 '13 at 3:19
  • $\begingroup$ This is a very similar problem to math.stackexchange.com/questions/342281/help-with-optimization You haven't answered my last comment on that one. Let's get that one first, then you might solve this one on your own. $\endgroup$ – Ross Millikan Mar 27 '13 at 3:19

Draw a picture.

Imagine that the upper left-hand corner is at $(x,x^{1/3})$. Then the height of the rectangle is $x^{1/3}$.

The width of the rectangle is $3-x$. (This won't make much sense without the picture.)

The area $A(x)$ of the rectangle is therefore given by $$A(x)=x^{1/3}(3-x).$$ Maximize $A(x)$, using the usual tools. Note that $0\le x\le 3$.

Added: We are maximizing $A(x)=3x^{1/3}-x^{4/3}$ over the interval $[0,3]$. We have $A'(x)=x^{-2/3}-\frac{4}{3}x^{1/3}$. We look for roots of $A'(x)=0$ in the interior of our interval. To solve, rewrite the equation as $x^{-2/3}=\frac{4}{3}x^{1/3}$ and multiply both sides by $x^{2/3}$. We get $x=\frac{3}{4}$. The endpoints $0$ and $3$ give $0$ area, so the maximum area is obtained by choosing $x=\frac{3}{4}$.

  • $\begingroup$ can you please show correct answer so i can follow your guidlines and correctness $\endgroup$ – Michael Rametta Mar 27 '13 at 3:28
  • $\begingroup$ I would really appreciate it $\endgroup$ – Michael Rametta Mar 27 '13 at 3:29
  • $\begingroup$ I will add it in at most a couple of hours. In the meantime, try to finish it, there really is only calculation to do. $\endgroup$ – André Nicolas Mar 27 '13 at 3:40
  • $\begingroup$ so would it be 1/3x^(-2/3)(3-x)+(x^1/3)(-1) $\endgroup$ – Michael Rametta Mar 27 '13 at 3:47
  • $\begingroup$ This user has also posted math.stackexchange.com/questions/340114/… so I suspect the problem is interpreting the words, not the math. $\endgroup$ – Ross Millikan Mar 27 '13 at 3:56

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