Find the absolute maximum and absolute minimum values of f on the given interval.
a) f(t)= t sqrt(36-t^2) [-1,6]
absolute max=
absolute min=
b) f(t)= 2 cos t + sin 2t [0,pi/2]
absolute max=
absolute min=

absolute max= (4.243,18)

absolute min =(-1,-5.916)

absolute max=(pi/6, 2.598)

absolute min = (pi/2,0)

Step-by-step explanation:

a)

To find max and minima in the given interval let us take log and differentiate

It is sufficient to find max or min of Y

In the given interval only 4.243 lies

And we find this is maximum hence maximum at  (4.243,18)

Minimum value is only when x = -1 i.e. -5.916

b)

Equate I derivative to 0

-2sint +1-2sin^2 t=0

sint = 1/2 only satisfies I quadrant.

So when t = pi/6 we have maximum

Minimum is absolute mini in the interval i.e. (pi/2,0)

518,400

step-by-step explanation:

the area is multiplying all the given measurements together.

40 * 18 * 30 * 24 = 518,400

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions