Differentiation problem

Question 1:
The circumference of the rectangle is 2x+2y=1000
and the area is xy
2x+2y=1000
x + y = 500
y = 500 – x
A = xy = x(500 – x)
A = -x^2 + 500x
Differentiate A with respect to x
dA/dx = -2x + 500
let dy/dx = 0
0 = -2x + 500
2x = 500
x = 250
y = 500 -250
y = 250
Take second derivative
d2A/dx2 = -2
negative value of d2A/dx2 indicates it hs a maximum
Thus x = 250 unit and y = 250 unit
The area is 250 8 250 = 62500 sq. unit

The maximum area is 62500 sq. unit

Question 2:

y=(t+1)^7(2t+3)^4(2t-1)^5,

d/dt (UVW) = UV (dW/dt) + UW (dV/dt) + VW (dU/dt)
Let U = (t+1)^7
dU/dt = 7(t +1)^6
Let V = (2t+3)^4
dV/dt = 2*4(2t + 3)^3 = 8(2t + 3)^3
Let W = (2t-1)^5
dW/dt = 2*5(2t -1)^4 = 10(2t - 1)^4
dy/dt = (t+1)^7(2t+3)^4 [10(2t - 1)^4] +(t+1)^7(2t-1)^5 [8(2t + 3)^3] + (2t+3)^4(2t-1)^5[7(t +1)^6]

dy/dt = 10(t+1)^7(2t+3)^4 (2t - 1)^4 +8(t+1)^7(2t-1)^5 (2t + 3)^3
+ 7(2t+3)^4(2t-1)^5(t +1)^6


2011-01-20 12:35:02 補充：
Question 2:
Using product rule for 3 terms. Each term is a polynomial function.
y = UVW
where U = (t+1)^7, V = (2t+3)^4, W = (2t-1)^5
Find dU/dt, dV/dt and dW/dt by chain rule

Q1
2x + 2y = 1000
x + y = 1000
y = 1000 - x

A = xy
= x (1000 - x)
= 1000x - x^2

dA/dx = 1000 - 2x
d2A/dx2 = -2 < 0

When dA/dx = 0,
1000 - 2x = 0
x = 500
y = 1000 - 500
= 500

When x = 500,
A attains its absolute maximum when x = 500
Max area = 500*500 = 250,000 units^2



Q2
y = (t+1)^7 (2t+3)^4 (2t-1)^5
dy/dx = (t+1)^7 (2t+3)^4 d[(2t-1)^5]/dt + (t+1)^7 d[(2t+3)^4]/dt (2t-1)^5 + d[(t+1)^7]/dt (2t+3)^4 (2t-1)^5
= (t+1)^7 (2t+3)^4 [5*2*(2t-1)^4] + (t+1)^7 [4*2*(2t+3)^3] (2t-1)^5 + [7(t+1)^6] (2t+3)^4 (2t-1)^5
= 10 (t+1)^7 (2t+3)^4 (2t-1)^4 + 8 (t+1)^7 (2t+3)^3 (2t-1)^5 + 7(t+1)^6 (2t+3)^4 (2t-1)^5