Dimension area problem

**funnytim** · October 31st 2009, 08:54 PM

Hey Guys!

I have a calculus problem here :

A book with cover dimensions 5in * 7in is to be placed symmetrically in a diamond-shaped gift box. What's the smallest possible area of the box? Also show how the answer is minimal.

Other than finding the area of the book (35in2) I am not sure how to approach this problem.

Thanks in advance for your assistance!

**davismj** · October 31st 2009, 09:40 PM

Well, I'm looking at this problem and noticing that in putting the book into the diamond box, the area of the diamond is equal to the area of the book plus the area of the four triangles created on each side of the book. So what we're looking to do is to minimize the size of the triangles, right?

**funnytim** · November 1st 2009, 12:25 AM

Ya, we're looking at something as attached.

On the four triangles at each end, I am given the base (5 and 7) but how do I calculate the "imaginary" height to be able to find the area of the triangles?

And then, how can I figure out how to minimize its area?

Thanks!

Edit: I just discovered that I can find the length of the sides of the diamond-shape (I think) by drawing a diagonal line across the book, then computing 5(squared) + 7(squared) to obtain (square root)74. (diagonal line across the book is parallel to the lengths of the diamond-shape, therefore they should be the same?)
Is it right to take (square root)74/2 = (square root)37 to be the length of the side of the 4 triangles, and then using pythagors to get (square root)5.92 to be the height of the triangle? (somehow this doesn't seem right though...)

**calum** · November 1st 2009, 12:39 AM

Originally Posted by funnytim

Ya, we're looking at something as attached.

On the four triangles at each end, I am given the base (5 and 7) but how do I calculate the "imaginary" height to be able to find the area of the triangles?

And then, how can I figure out how to minimize its area?

You will notice that the two angles between the corner of the book and the box adds to 90 degrees.

The top triangle's area is given by: $A = \frac{7}{2} \times \frac{7}{2}tan \theta$

The side triangle's area is giveb by: $A = \frac{5}{2} \times \frac{5}{2}tan (\frac{\pi}{2} - \theta)$

So the total area of the triangles is given: $A = \frac{49}{2}tan \theta + \frac{25}{2}tan(\frac{\pi}{2} - \theta)$
Differentiate this equation and then solve the derivative for theta to find the minimum value of theta.

**funnytim** · November 1st 2009, 12:48 AM

Would you mind explaining how you got $<br /> A = \frac{7}{2} \times \frac{7}{2}tan \theta<br />$
and $A = \frac{5}{2} \times \frac{5}{2}tan (\frac{\pi}{2} - \theta$ ?Thanks again

**calum** · November 1st 2009, 01:00 AM

Originally Posted by funnytim

Would you mind explaining how you got $<br /> A = \frac{7}{2} \times \frac{7}{2}tan \theta<br />$
and $A = \frac{5}{2} \times \frac{5}{2}tan (\frac{\pi}{2} - \theta$ ?Thanks again

Draw a vertical line down from the top peak of the diamond to the book. This line should be perpendicular to the line of the book.

Call the angle in the bottom left of this new triangle (where the book corner touches the box) theta.

You should have a right-angled triangle with base length 7/2 and corner angle theta.

The height of this triangle is now 7/2 * tan(theta). The area of the triangle is half the base times the height, but there are two of these triangles that make up the larger top triangle. So the total area is base*height.

**funnytim** · November 1st 2009, 01:30 AM

Thanks a lot for explaining that, makes that part clear now

Solving for the derivative of the total area I got:

49/2 sec[squared] theta + 25/2 sec[squared] theta ([pie]/2 - theta)

Is that correct? If so how can I then solve for theta.

I tried factoring out the sec[squared] theta, but that didn't really help...:S

Thanks again, I'm pretty bad with calculus :P

**calum** · November 1st 2009, 01:31 AM

Originally Posted by funnytim

Thanks a lot for explaining that, makes that part clear now

Solving for the derivative of the total area I got:

49/2 sec[squared] theta + 25/2 sec[squared] theta ([pie]/2 - theta)

Is that correct? If so how can I then solve for theta.

I tried factoring out the sec[squared] theta, but that didn't really help...:S

Thanks again, I'm pretty bad with calculus :P

$\frac{dA}{d \theta} = \frac{49}{2}sec^2 \theta - \frac{25}{2}sec^2(\frac{\pi}{2} - \theta)$

Solving:
-multiply by 2 and use sec = 1/cos
$0 = \frac{49}{cos^2 \theta} - \frac{25}{cos^2(\frac{\pi}{2} - \theta)}$

Now use the complimentary angle rule ( $cos(\frac{\pi}{2} - \theta) = sin \theta$ ):

$0 = \frac{49}{cos^2 \theta} - \frac{25}{sin^2 \theta} = \frac{49sin^2 \theta - 25cos^2 \theta}{cos^2 \theta sin^2 \theta}$

Multiplying out the denominator and using the Pythagorean identity:

$0 = 49sin^2 \theta - 25cos^2 \theta = 49 - 49cos^2 \theta - 25cos^2 \theta = 49 - 74cos^2 \theta$

$cos \theta = \sqrt{\frac{49}{74}}$

$\theta = cos^{-1}(\sqrt{\frac{49}{74}})$

**EvaMB** · November 1st 2009, 02:09 PM

Would you be able to explain why it is "(tan (pi/2 - theta))" for the second triangle? Thanks.

**funnytim** · November 4th 2009, 09:18 PM

Thanks calum!

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