i have calculus questions...anyone wanna help, please?!?

Question

i was wondering if anyone could help me with explaining how to do some of these problems..thanks soo much!



1. When is the function concave up, if f'(x)=(x-2)^2(x-3)^2

2.Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the parabola y=16-5x^2

3. A point moves on the curve 6x^2-3y^2=18 so that its x-coordinate increases at the constant rate of 9 m/s. At what rate is the y-coordinate chaing when y=3 meters.

4.The acceleration of a particle moving along the coordinate line is 5+4t m/s^2. At t=0 the velocity is 5 m/s. find how far the particle moves from time t=0 to t=2.


5. A container with rectangular base, rectangular sides, and no top is to have a volume of 16 m^3. The width of the base is to be 2 m. When cut to size, material costs $27 per square meter for the base and $9 per square meter for the sides. What is the cost of the least expensive container?



THANKS FOR ANY OF YOUR HELP!

Answer 1

I can help with #'s 1 and 3.

1. A function is concave up when the second derivitive, f''(x), is positive. You need to take the derivitive of f'(x) to get f''(x), and then figure out what values of x make f''(x) positive.

3. Use the chain rule to solve this:

dy/dt = (dx/dt)*(dy/dx)

We are told that (dx/dt) is 9 m/s. Figure out what (dy/dx) is when y=3, and you will solve the problem.

Hope this helps.

Answer 2

Answer to 4:

s = ut + (1/2)at^2
s = 5(2) + (1/2)(5+4(2))(2^2)
s = 10 + (1/2)(12)(4)
s = 10 + 24
s = 34 m