Alrighty
squaer base so length=width, nice
v=lwh
but in this case, l=w, so replace l with w
V=w²h
and volume is 32000
32000=w²h
the amount of materials is the surface area
note that there is no top
so
SA=LW+2H(L+W)
L=W so
SA=W²+2H(2W)
SA=W²+4HW
alrighty
we gots
SA=W²+4HW and
32000=W²H
we want to minimize the square foottage
get rid of one of the variables
32000=W²H
solve for H
32000/W²=H
subsitute
SA=W²+4WH
SA=W²+4W(32000/W²)
SA=W²+128000/W
take derivitive to find the minimum
dSA/dW=2W-128000/W²
where does it equal 0?
0=2W-1280000/W²
128000/W²=2W
128000=2W³
64000=W³
40=W
so sub back
32000/W²=H
32000/(40)²=H
32000/(1600)=H
20=H
the box is 20cm height and the width and length are 40cm
Answer:
f/g at t =3 will be 9/8
d/ddx(f(x)/g(x))=f'(x)d(x)-f(x)g'(x)/(g'(x))^2
rate of change=(6*8)+(9*9)/9^2
rate of change =43/27=1.5925
(x+3 ; y-5)
If you have (-2 ; 3) that means x = -2 and y = 3 so:
(-2+3 ; 3-5) = (1 ; -2)
