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Math Help - min. area of triangle

  1. #1
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    min. area of triangle

    p228 q20c
    given area of triangle A = \frac{2(r^2+r+1)}{(1+r)^2}
    find the min. area
    i put r --> infinty then A = 2 but it is wrong. the answer of my book says A=3/2

    thanks in advance
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    p228 q20c
    given area of triangle A = \frac{2(r^2+r+1)}{(1+r)^2}
    find the min. area
    i put r --> infinty then A = 2 but it is wrong. the answer of my book says A=3/2

    thanks in advance
    A = 2\left(1 - \frac{r}{(1+r)^2}\right)

    A achieves minimum when \frac{r}{(1+r)^2} is maximum.

    But (1+r)^2 \geq 4r \Rightarrow \frac{r}{(1+r)^2} \leq \frac14 \Rightarrow \left(1 - \frac{r}{(1+r)^2}\right) \geq \frac34 \Rightarrow A \geq \frac32

    With equality at r=1.

    So minimum of A is achieved at r=1.And that is A = 3/2.
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  3. #3
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    how to prove
    (1+r)^2 \geq 4r
    if it is (k+r)^2 , then (k+r)^2 \ge ?
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  4. #4
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    Quote Originally Posted by afeasfaerw23231233 View Post
    how to prove
    (1+r)^2 \geq 4r
    [snip]
    To prove that 1 + 2r + r^2 \geq 4r you only have to prove that 1 + r^2 \geq 2r.

    Draw a graph of y = x^2 + 1 and y = 2x and it's not hard to see (and then prove) that the line y = 2x is tangent to y = x^2 + 1 at (1, 2) ........
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    [tex][snip]when \frac{r}{(1+r)^2} is maximum.

    [snip]at r=1.
    [snip]
    This result can also be easily proved using calculus.
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  6. #6
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    @mr fantastic: 1+r^2\ge 2r \iff r^2-2r+1\ge 0 \iff \left(r-1\right)^2\ge 0 \,\Box.
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