Problem of maximization

有沒有規定用甚麼方法?

2011-01-19 21:33:30 補充：
x^2/16 + y^2/9 = 1

y = 3√(1 - x^2/16)

Let the area of the rectangle is A. Then

A

= 4xy

= 4x{3√(1 - x^2/16)}

= 12x√(1 - x^2/16)

= 3x√(16 - x^2)

Now, maximize A is equal to maximize A^2 = 9x^2(16 - x^2)

(A^2)' = 9x^2(-2x) + (16 - x^2)(18x) = -36x^3 + 288x

Set (A^2)' = 0 =>x = -2√2 or 2√2 and then y = -3/√2 or 3/√2

By, physical consideration or second derivative test, x = -2√2 or 2√2 should give the maximum value of A.

The area of the largest rectangle

= 4xy

= 4(2√2)(3/√2)

= 24

just using differentiation

2011-01-19 21:03:24 補充：
By the way,
in this question,
http://hk.knowledge.yahoo.com/question/question?qid=7011011900433
y u have to sub dx/dt = -4 instead of 4?