A=x(12-x^2),
A'=12-3x^2
Finding critical value,u find x=2
Then A(2)=16
Thus (2,16) is a critical point and
A''(2)=-8<0 therefore (2,16) is relative maximum point,
Solving for y from y=12-x^2 we get 8
Therefore;length (x)=2 & width(y)=8
Area= 16 units squared
Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y=4−x^2.
Width =
Height =
2 answers
Why did it take 5 years to answer this question?