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Math Help - Urgent Optimization Questions

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by Super Mallow View Post
    ...

    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.
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  3. #3
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    Quote Originally Posted by Super Mallow View Post
    ...

    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
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  4. #4
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    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!
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  5. #5
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    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...
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