Shin C. answered • 04/09/20
UCLA Alumni | AP Calculus AB/BC & College Calculus Specialist
Hello Anna! To start answering this question, recall that the area of the triangle is A = 0.5xy (let x be the horizontal length, and y be the vertical length).
In right triangles, thanks to the Pythagorean theorem, x and y can be related such that x^2 + y^2 = 3^2. Rearrange that so that y is in terms of x: y = sqrt( 9 - x^2 ). Plug that back to the area formula to get
A = 0.5x * sqrt(9 - x^2). Now, in order to find the maximum area, you need to take the derivative of A(x) and find x values so that A ' (x) = 0 or DNE (does not exist). IF, at that x value, A'(x) changes from positive to negative, then that indicates a relative maximum.
A ' (x) = ( sqrt( 9 - x^2 ) / 2 ) - ( x^2 / ( 2 * sqrt( 9 - x^2 ) ) ).
A'(x) = 0 when x = 3sqrt(2) / 2, and checking signs does indeed indicate that A'(x) changes from positive to negative.
A'(x) DNE when x = 3 (because in the second term, you cannot have a number divided by sqrt of 0, right?), and because you can't put any number higher than x = 3, only A ( 3 sqrt2 / 2 ) yields the maximum area.
A (3 sqrt(2) / 2) = 0.5( 3sqrt2 / 2) * sqrt(9 / 2) = 2.25...................(ANS)
Hope this helps! Let me know in the comments if additional assistance/questions arise!
