Author Topic: Heinemann question (Read 1229 times) Tweet Share

Boots · « **on:** April 05, 2010, 09:01:14 am »

Can someone please help with this question:

A closed box has a total surface area of 500cm sq. If the box is to have square ends, find the greatest volume possible.

the.watchman · « **Reply #1 on:** April 05, 2010, 09:49:28 am »

Let the box have dimensions p x p x q (the two equal dimensions gives it square ends)

(TSA = Total surface area AND V(p) = Volume)

Therefore $TSA = 2p^2 + 4pq = 500$

$4pq = 500-2p^2$

$q = \frac{250-p^2}{2p}$

Hence, $V(p) = p^2 \cdot q = p^2 \cdot \frac{250-p^2}{2p} = -\frac{1}{2}p(p^2-250)$

Because $V'(p) = 0$ for stationary points

AND $p \in (0,5\sqrt{10})$

So $V'(p) = 125 - \frac{3p^2}{2} = 0$

$p=\frac{5 \sqrt{30}}{3}$ ( $p=-\frac{5 \sqrt{30}}{3}$ IS REJECTED AS $p \in (0,5\sqrt{10})$ )

SO the greatest possible volume is given by $V(6) = \frac{1250\sqrt{30}}{9} = 760.726$ (to 3dp)

Please let me know if I've stuffed it up!

brightsky · « **Reply #2 on:** April 05, 2010, 10:01:39 am »

Let the sides of the square be $x$ cm and the length of the box be $y$ cm.

Then $2(x^2 + xy + xy) = 500$

$2(x^2 + 2xy) = 500$

$x^2 + 2xy = 250$

$y = \frac{250 - x^2}{2x}$ ...(1)

Now the volume is given by $x^2y$ ...(2)

Substitute (1) into (2):

$x^2 \cdot \frac{250 - x^2}{2x} = \frac{x(250 - x^2)}{2}$

Consider the graph $y = \frac{x(250 - x^2)}{2}$ .

The local maximum would be at a stationary point, hence we need to find when $\{x:\frac{dy}{dx} = 0\}$ .

$\frac{dy}{dx} = \frac{250 - x^2}{2} - x^2 = 0$

Solving that gives $x = \pm 5 \sqrt{\frac{10}{3}}$ , but, obviously $x > 0$ so $x = 5 \sqrt{\frac{10}{3}}$ .

Substituting that back into (1):

$y = 5 \sqrt{\frac{10}{3}}$

Hence the largest volume is $x^2y = \left(5 \sqrt{\frac{10}{3}}\right)^3 = \frac{1250\sqrt{\frac{10}{3}}}{3} \approx 760.73$ .

Hope I did this correctly..

EDIT: Beaten! xD

the.watchman · « **Reply #3 on:** April 05, 2010, 10:05:55 am »

Lol, be careful not to double define variables, may void solution

Boots · « **Reply #4 on:** April 05, 2010, 10:15:47 am »

Thank you. I think I just misunderstood what square ends are.

What about this question:
Joseph is standing on a long narrow point of land which juts 5km out to the sea directly opposite the surf club. The finish line is 8km along the shore from the surf club. If joseph can swim at 7km/h and run at 16km/h, find the position along the shore that he should aim for in order to minimise the time taken to complete the race.

If you could just simply provide the equation to solve, or the diagram, I will do the rest. Thanks

Boots · « **Reply #5 on:** April 05, 2010, 10:16:20 am »

Quote from: brightsky on April 05, 2010, 10:01:39 am

Let the sides of the square be $x$ cm and the length of the box be $y$ cm.

Then $2(x^2 + xy + xy) = 500$

$2(x^2 + 2xy) = 500$

$x^2 + 2xy = 250$

$y = \frac{250 - x^2}{2x}$ ...(1)

Now the volume is given by $x^2y$ ...(2)

Substitute (1) into (2):

$x^2 \cdot \frac{250 - x^2}{2x} = \frac{x(250 - x^2)}{2}$

Consider the graph $y = \frac{x(250 - x^2)}{2}$ .

The local maximum would be at a stationary point, hence we need to find when ${x:\frac{dy}{dx} = 0}$ .

$\frac{dy}{dx} = \frac{250 - x^2}{2} - x^2 = 0$

Solving that gives $x = \pm 5 \sqrt{\frac{10}{3}}$ , but, obviously $x > 0$ so $x = 5 \sqrt{\frac{10}{3}}$ .

Substituting that back into (1):

$y = 5 \sqrt{\frac{10}{3}}$

Hence the largest volume is $x^2y = \left(5 \sqrt{\frac{10}{3}}\right)^3 = \frac{1250\sqrt{\frac{10}{3}}}{3} \approx 760.73$ .

Hope I did this correctly..

EDIT: Beaten! xD

lol. Thanks anyway.

brightsky · « **Reply #6 on:** April 05, 2010, 10:28:09 am »

Quote from: the.watchman on April 05, 2010, 10:05:55 am

Lol, be careful not to double define variables, may void solution

True..but the GRAPH $y = ....$ is a special case...is it not?

the.watchman · « **Reply #7 on:** April 05, 2010, 10:37:52 am »

Quote from: brightsky on April 05, 2010, 10:28:09 am

Quote from: the.watchman on April 05, 2010, 10:05:55 am
Lol, be careful not to double define variables, may void solution

True..but the GRAPH $y = ....$ is a special case...is it not?

Not really, in the case of graphs, use $y_1$ and $y_2$ ETC.
It is safer to go with V for volume

Boots · « **Reply #8 on:** April 05, 2010, 10:58:27 am »

dnt forget my other question!!!

brightsky · « **Reply #9 on:** April 05, 2010, 11:06:37 am »

Can't seem to upload my diagram, but just for a rough sketch of my interpretation:

\ Joseph /
\ /
V
|
| 5km
WATER | WATER
|
___________________________________________________________________________

SHORE Surf Club <------------------------------------>Finish Line
8km

EDIT: Jacob changed to Joseph. :p

brightsky · « **Reply #10 on:** April 05, 2010, 11:13:49 am »

Let the km horizontally away from the surf club (indicating where the position he needs to aim for on the shoreline) be $x$ km.

Let $d_1$ be the distance he swims and $d_2$ be the distance he runs.

Hence, $d_1 = \sqrt{x^2 + 5^2} = \sqrt{x^2 + 25}, x \in [0, 8]$ ...(2)

$d_2 = 8 - x$ ...(2)

$\because$ The total time taken by the man to complete the race is given by: $\frac{d_1}{7} + \frac{d_2}{16}$

$\therefore$ The total time is $\frac{\sqrt{x^2 + 25}}{7} + \frac{8 - x}{16}$

(Differentiate that and let the derivative equal to 0 to find the minimum time.)

-Stop reading where you want to do the rest yourself (mine might be wrong anyway :p)-

Consider the graph $f(x) = \frac{\sqrt{x^2 + 25}}{7} + \frac{8 - x}{16}$

$f'(x) = \frac{x}{7\sqrt{x^2 + 25}} - \frac{1}{16}$

Let $f'(x) = 0$ to find the local maximum.

Solving that gives $x = \frac{35}{3\sqrt{23}} \approx 2.433$ .

Hence, the point the man wants to aim for is $\frac{35}{3\sqrt{23}}$ km horizontally away from the surf club. That is to say, he swims to that point, and then runs horizontally for the remaining kilometres to the finish line.

brightsky · « **Reply #11 on:** April 05, 2010, 11:31:23 am »

For verification that it is the local minimum, the graph of $f(x) = \frac{\sqrt{x^2+25}}{7} + \frac{8-x}{16}$ looks like this.

Boots · « **Reply #12 on:** April 05, 2010, 12:16:49 pm »

Wow amazing thank you sooo much.
Btw do you like twillight? Cos' you interpreted Joseph as Jacob

Boots · « **Reply #13 on:** April 05, 2010, 12:27:03 pm »

One last question guys..please.

Find the largest area of a rectangular window that can be placed in the isoceles triangular cross-section of a roof as shown in attachment.

brightsky · « **Reply #14 on:** April 05, 2010, 02:51:33 pm »

Wow, interesting question...

Ok, let the length of the vertical side of the rectangle be $x$ units and the length of the horizontal side be $y$ units.

To avoid confusion, label the triangle as $\triangle ABC$ with A being the top of the triangle (according to the images provided, B being the left hand corner, and C being the right hand corner.

Let $\angle ACB = \theta$ .

Then $\angle ABC = \angle BAC = \frac{180^{\circ} - \theta}{2}$

Label the four corners, starting from the top left hand corner going clockwise, $DEFG$

$\angle FEC = 90^{\circ} - \theta \implies \angle AED = \theta \implies \angle ADE = \frac{180^{\circ} - \theta}{2}$

So $\triangle ABC \sim \triangle ADE$ (AAA)

Hence $AE = y$ , so $EC = 2 - y$

So in $\triangle ECF, \sin \theta = \frac{x}{2 - y} \implies y = 2 - \frac{x}{\sin\theta}$

This $\theta$ is a bit annoying...think you need to turn in into an expression in terms of $x$ .

Not too sure how to do this, but one method is perhaps solving these three equations simultaneously:

$\frac{x}{\sin\left(\frac{180^{\circ} - \theta}{2}\right)} = \frac{BG}{\sin\left(\frac{\theta}{2}\right)}$ (found in $\triangle DGB$ )...(1)

$\frac{FC}{\sin(90^{\circ} - \theta)} = \frac{x}{\sin\theta}$ ...(2)

$BG + FC + \left(2 - \frac{x}{\sin\theta}\right) = 2$ ...(3)

When you find $\theta$ , simply sub it into the expression for the area of the rectangle:

$xy = x \cdot \left(2 - \frac{x}{\sin\theta}\right)$ and then differentiate it, let the derivative equal to 0, and then solve for x, solve for y, and hence get the area.

Hope this helps and I hope that it's not devastatingly wrong...as it appears to be..:p

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