1. ## Urgent Optimization Questions

Yes, I got 2 more problems....I really need some solutions to these...

1) You're desgning a rectangle poster to contain 50 in squared of printing with a 4-inch margin at the top and bottom and a 2-inch margin at each side. What overall dimensions will minimize the amount of paper used?

2) Two sides of a trianglehave lengths A and B, and the angle between them is theta. What value of theta will maximize the triang;e's area? Hint: A=1/2ABSinTheta

I'm in calculus and I do not know how to solve these problems. Any help is appreciated!

2. Originally Posted by Super Mallow
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1) You're desgning a rectangle poster to contain 50 in squared of printing with a 4-inch margin at the top and bottom and a 2-inch margin at each side. What overall dimensions will minimize the amount of paper used?

...
Hello,

let x be the width and y the length of the printing area. Then the size of the poster is:

$A=(x+4)(y+8) = xy+8x+4y+32$

You know that $x \cdot y = 50 ~\iff~y=\frac{50}{x}$ . Plug in these values into the equation of A:

$A(x) = 50 + 8x + \frac{200}{x} + 32$ . You get the extreme (minimum or maximum) value if the 1st derivative of A equals zero:

$A'(x) = 8-\frac{200}{x^2}$ . Solve the equation

$A'(x) = 0$ for x. I've got $x = \frac52 \cdot \sqrt{2}$ . Use this value to calculate y. I've got $y = 10 \cdot \sqrt{2}$

And now you can calculate the length and the width of the complete poster.

3. Originally Posted by Super Mallow
...

2) Two sides of a trianglehave lengths A and B, and the angle between them is theta. What value of theta will maximize the triang;e's area? Hint: A=1/2ABSinTheta
Hi,

take the equation of the function:

$f(\theta)=\frac12 \cdot A \cdot B \cdot \sin(\theta)$

You'll get the extreme value if $f'(\theta)=0$:

$f'(\theta)=\frac12 \cdot A \cdot B \cdot \cos(\theta)$

Solve $f'(\theta)=0$ for $\theta$ . I've got $\theta = 90^\circ$

4. Huge thanks man! I still don't get the first one..the answer in the book is 9x18, I don't see how to get tht answer!

5. I got that x=5. that makes y=10. When you plug that into the min area equation, you'll get (5+4)(10+8) which will give you your min dimensions.

And yes, I know this post is moldy oldy, but it may help the few people who use the search function...