Question 119021
You have two equations:

1) {{{b+h=25}}}


2) {{{A=bh/2}}}


And we want to maximize A.


From 1) we can say:  {{{h=25-b}}} and with this information we can re-write 2)


3) {{{A=b(25-b)=25b-b^2}}}


{{{A=-b^2+25b+0}}} describes a concave down ({{{a<0}}}) parabola with vertex at {{{b=-25/(2*(-1))=12.5}}}.  Since this is a concave down parabola, we know that the vertex point is a maximum.


(Alternatively, if you are taking calculus at this point, you can set the first derivative equal to zero and solve for x to find the x coordinate of the local minimum/maximum, and then evaluate the sign of the second derivative to determine that the point is a maximum, thus:

{{{dA/db=25-2b}}}
{{{25-2b=0}}}
{{{b=12.5}}}


{{{d^2(A)/db^2=-2<0}}}, therefore the point is a maximum)


We have determined the length of the base that produces the maximum volume, so we can calculate the height by {{{h=25-12.5=12.5}}}.  And finally, the maximum Area by {{{A[max]=(12.5^2)/2=156.25/2=78.125cm^2}}}.


You should actually state your answer as {{{78cm^2}}} because your answer should never be stated with more precision than the least precise measurement given in the problem.  Since the sum of the base and height were given to the nearest whole centimeter, your answer should also be to the nearest whole centimeter.