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
It would be 5%. I know the answer is right I just don’t know how to explain it. Sry. But I hope this helps.
Answer:
36 and 48 and 96
Step-by-step explanation:
add 3+4+8=15
all of the angles in a triangle have to add up to 180
so 180÷15=12
now for this ratio, each part will equal 12
so we have to multiply each of the parts by 12
3×12=36
4×12=48
8×12=96
those are the measures of the angles
you can add them up again and see that they equal 180
you can try dividing and you will see that they also equal the ratio
Answer:
1595 ft^2
Step-by-step explanation:
The answer is obtained by adding the areas of sectors of several circles.
1. Think of the rope being vertical going up from the corner where it is tied. It goes up along the 10-ft side. Now think of the length of the rope being a radius of a circle, rotate it counterclockwise until it is horizontal and is on top of the bottom 20-ft side. That area is 3/4 of a circle of radius 24.5 ft.
2. With the rope in this position, along the bottom 20-ft side, 4.5 ft of the rope stick out the right side of the barn. That amount if rope allows for a 1/4 circle of 4.5-ft radius on the right side of the barn.
3. With the rope in the position of 1. above, vertical and along the 10-ft left side, 14.5 ft of rope extend past the barn's 10-ft left wall. That extra 14.5 ft of rope are now the radius of a 1/4 circle along the upper 20-ft wall.
The area is the sum of the areas described above in numbers 1., 2., and 3.
total area = area 1 + area 2 + area 3
area of circle = (pi)r^2
total area = 3/4 * (pi)(24.5 ft)^2 + 1/4 * (pi)(4.5 ft)^2 + 1/4 * (pi)(14.5 ft)^2
total area = 1414.31 ft^2 + 15.90 ft^2 + 165.13 ft^2
total area = 1595.34 ft^2
