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:
perpindicular slope = -2/3
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Substitute the following values into the cylindrical volume equation below:
r= 30/2 = 15 cm
h = 320 cm
V=πr2h
V=(3.14)(15)^2 (320)
V=226 080 cm^3
Answer:
23.28 inches (im not sure if this is correct because you need length, height, and width.)
Step-by-step explanation:
<u>lwh</u>
3
Answer:
Option A. 
Step-by-step explanation:
we know that
In the right triangle MNL
Applying the Pythagoras Theorem
substitute the values
