Given that <span>Line m is parallel to line n.
We prove that 1 is supplementary to 3 as follows:
![\begin{tabular} {|c|c|} Statement&Reason\\[1ex] Line m is parallel to line n&Given\\ \angle1\cong\angle2&Corresponding angles\\ m\angle1=m\angle2&Deifinition of Congruent angles\\ \angle2\ and\ \angle3\ form\ a\ linear\ pair&Adjacent angles on a straight line\\ \angle2\ is\ supplementary\ to\ \angle3&Deifinition of linear pair\\ m\angle2+m\angle3=180^o&Deifinition of supplementary \angle s\\ m\angle1+m\angle3=180^o&Substitution Property \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7C%7D%0AStatement%26Reason%5C%5C%5B1ex%5D%0ALine%20m%20is%20parallel%20to%20line%20n%26Given%5C%5C%0A%5Cangle1%5Ccong%5Cangle2%26Corresponding%20angles%5C%5C%0Am%5Cangle1%3Dm%5Cangle2%26Deifinition%20of%20Congruent%20angles%5C%5C%0A%5Cangle2%5C%20and%5C%20%5Cangle3%5C%20form%5C%20a%5C%20linear%5C%20pair%26Adjacent%20angles%20on%20a%20straight%20line%5C%5C%0A%5Cangle2%5C%20is%5C%20supplementary%5C%20to%5C%20%5Cangle3%26Deifinition%20of%20linear%20pair%5C%5C%0Am%5Cangle2%2Bm%5Cangle3%3D180%5Eo%26Deifinition%20of%20supplementary%20%5Cangle%20s%5C%5C%0Am%5Cangle1%2Bm%5Cangle3%3D180%5Eo%26Substitution%20Property%0A%5Cend%7Btabular%7D)
</span>
We prove that 1 is supplementary to 3 as follows:
Mark wants to know how many families in his neighborhood plan to attend the neighborhood party. He puts all 80 of the neighborho
2 answers:
You might be interested in
The Lagrangian
has critical points where the first derivatives vanish:
We can't have , since that contradicts the last condition.
(0 critical points)
If two of them are zero, then the remaining variable has two possible values of . For example, if
, then
.
(6 critical points; 2 for each non-zero variable)
If only one of them is zero, then the squares of the remaining variables are equal and we would find (taking the negative root because
must be non-negative), and we can immediately find the critical points from there. For example, if
, then
. If both
are non-zero, then
, and
and for either choice of , we can independently choose from
.
(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)
If none of the variables are zero, then . We have
and similary have the same solutions whose signs can be picked independently of one another.
(8 critical points)
Now evaluate at each critical point; you should end up with a maximum value of
and a minimum value of
(both occurring at various critical points).
Here's a comprehensive list of all the critical points we found: