Question

Find the minimum distance from the parabola
x + y^2 = 0
to the point (0,-3).

Answer 1

The manager of a large apartment complex knows from experience that 90 units will be occupied if the rent is 360 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 9 dollar increase in rent. Similarly, one additional unit will be occupied for each 9 dollar decrease in rent. What rent should the manager charge to maximize revenue?

Answer 2

A particle is moving with acceleration a(t) = 30 t + 12. its position at time t =0 is s(0) = 11 and its velocity at time t =0 is v(0) = 16. What is its position at time t = 8?

Answer 3

the square of the distance from a point (x,y) to the point (0,3) is given by
d^2 = [(x-0)^2 + (y+3)^2]
= x^2 +y^2 + 6 y + 9

in this case x = -y^2
so
d^2 = z = y^4 + y^2 + 6 y + 9
find dz/dy and set it to 0
0 = 4 y^3 +2 y + 6
hmm, try y = -1
0 - -4 -2 + 6 sure enough

Answer 4

d units/d rent = -1/9 = du/dr
du = -(1/9) dr
u = -(1/9) r + C
when r = 360, u = 90
90 = -(1/9)360 + c
810 = -360 + 9c
c = 130
so
u = -(1/9)r + 130
revenue = z = u r
z = u r = -(1/9)r^2 + 130 r
dz/dr = 0 = -(2/9)r + 130
r = 585

Answer 5

a(t) = 30 t + 12
v = Vi + 15 t^2 + 12 t
x = Xi + 5 t^3 + 6 t^2
now put in your initial conditions to find Vi and Xi and therefore x(t)