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Show that the area ,A,of the rectangle PQRS is given by $2p(16-p^2)$ enter image description here For question 10a) I am having trouble finding the $p^2$ I understand the 2p and is able to find the 16 using dy dx=0 Please help (https://i.stack.imgur.com/GemEv.jpg)

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  • $\begingroup$ Note: $S=4-p, R=4+p$. Substitute to $y$ and then area. $\endgroup$ – farruhota May 11 '18 at 3:44
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  1. The area of the rectangle $PQRS$ is the length of $RS$ multiplied by the length of $PS$. You are given the length of $RS$ as being $2p$. You want to find the length of $PS$.

  2. Let $(x,y)$ be the coordinates of $P$. Since $P$ is on the graph, we must have that $y = x(8-x)$. So what we want to find is the $x$-coordinate of $P$.

  3. Note that $S$ has the same $x$-coordinate as $P$. So if we can find the $x$-coordinate of $S$ then we can do step (2) and be finished.

  4. To get the $x$-coordinate of $S$, note that the rectangle is symmetric and that the middle of the rectangle is aligned with the middle of the graph. Indeed the $x$-coordinate is distance $p$ to the left of the middle of the graph. Hence the $x$-coordinate of $S$ must be at $x = 4-p$.

  5. Plugging in $x = 4 - p$ into $y = x(8-x)$ gives $y = (4-p)(4+p) = 16 - p^2$. Hence $A = 2p(16 - p^2)$.

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  • $\begingroup$ You're welcome @bryan $\endgroup$ – Patrick Hew May 11 '18 at 5:35
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Let $S(4-p,0)$, then $P(4-p,(4-p)(4+p))$ or $P(4-p,16-p^2)$. Then you have $A=SR\times SP$.

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The picture is symmetrical about $x=4$, hence the $x$-coordinate of $P$ is $4-p$, hence the height is $(4-p)(8-(4-p))=(4-p)(4+p)= 16-p^2$

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