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Having some problems with these questions:

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Q3b) What's the best way to find the domain? I was initially pretty lost but would considering $x - 12>0$ and $y-20>0$ be best? Also I'm not sure why the answer has the domain as exclusive $12$ but inclusive $60$. Shouldn't they both be exclusive?

Q4b) Why don't we need to state the domain of $V$? The answer only gave the domain of $x$.

Q6d) how do we determine the shape? Area = $6a/(a+2)$.

Thanks

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    It is a bit long for a single question. Maybe you should split it into 3 questions. Moreover, as this is clearly a homework, it is usually appreciated if you show your thoughts on the question. Finally, note that this question has not much to do with "principal-ideal-domains"... – 2017-02-16
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    "Also I'm not sure why the answer has the domain as exclusive 12 but inclusive 60. Shouldn't they both be exclusive?" No. Because 0 < Area $\le$ 160. The area must be greater than 0 which means x > 12. And the area *can* be as large as 160 which means x *can* be as large as 60. – 2017-02-16
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    "but would considering x−12>0 and y−20>0 be best" If the function input is $x$ and the function is terms of x only and not y, then you only need to find the possible values of x. x -12 > 0 or x > 12 is part of it. But you also must have x $\le$ 60. – 2017-02-16
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    To figure out the appropriate domain in 3b) you must first do 3a) and find the formula for the area. I will not do that for you. But if you tell me what you have I will explain how to find the domain. – 2017-02-16
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    "Why don't we need to state the domain of V? The answer only gave the domain of x." V is a fixed constant. It isn't a variable and it isn't input. – 2017-02-16
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    6d) The rectangle is just the points O = (0,0). B = (0, f(a)), C = (a, f(a)), D = (a, 0). So the area is $a\times f(a) $. – 2017-02-16
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    Thanks fleablood. For 6d, I found the equation but I'm unsure on how to determine its shape? – 2017-02-16

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