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Thread: Surface defined on the domain?

  1. #1
    Junior Member
    Mar 2011

    Surface defined on the domain?

    "Let S be the surface z = e^(x+y) - 2, defined on the domain D = {(x,y) ∈ ℝ^2 | x ≥ 0 and y ≥ 0}. What is the minimal value of z for points on S?"

    With this question, I thought that since both the x- and y-values need to be greater than zero for the domain, then obtaining the minimal value would simply mean subbing 0 into the equation. I did that, and got the answer, -1. (Which also matches up as being the correct answer to the question). So, I just basically want to know was what I did to get the answer correct or is there some other way you get the answer? (As, I may be wrong but it just seems that it was a bit of a straightforward and simple approach with getting the answer.)
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  2. #2
    Senior Member abhishekkgp's Avatar
    Jan 2011
    what you did is correct. it can be explained as follows:
    1) the value of z completely depends on the value of x+y.
    2)the lower the value of x+y the lower the value of z.
    3) the least value of x+y available in D is 0.
    so e^0-2 is the answer.

    Here the situation was fairly simple so we could use the above reasoning. In general such problems are solved using multivariate calculus.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Nov 2010
    Madrid, Spain
    We have

    $\displaystyle -1\leq e^{x+y}-2\Leftrightarrow 1\leq e^{x+y}$

    The second inequality is true for all $\displaystyle x\geq 0,y\geq 0$ so, $\displaystyle -1$ is a lower bound for

    $\displaystyle z=f(x,y)=e^{x+y}-2$

    On the other hand we get that bound: $\displaystyle f(0,0)=-1$ .

    Edited: Sorry, Ididn't see the previous post.
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