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Andreas93 [3]
3 years ago
8

If 4,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the

box. Step 1 Let b be the length of the base and h the height. Then we must maximize the volume of the box, V
Physics
2 answers:
Sloan [31]3 years ago
5 0

Answer:

32000cm^3

Explanation:

Length L be base length

h be the height

Volume=V

V=L^2h

Surface area if box= L^2+4lh=4800

Solving for h

(4800-l^2/4L)=h

1200L-L^3/4=h

Substitute for h in V =L^2h

v=1200L-L^3/4

Differenting =1200-3*L^2/4=0

L=√(1600)=40

Substitute L=40

In V=1200L-L^3/4

Hence 1200*40-40^4/4=32000cm^3

bagirrra123 [75]3 years ago
4 0

Answer:

V = 32000 cm^3

the maximum volume of the box, V is 32000cm^3

Explanation:

Given;

Total surface area of box

S.A = 4800cm^2

Volume of a box with square base;

V = (l^2)h .....1

Surface area of a box with square base and opened top;

S.A = l^2 + 4lh = 4800 .......2

To be able to maximize Volume V, we need to reduce the number of variables in the equation 1

From equation 2, making h the subject of formula;

h = (4800 - l^2)/4l

Substituting h into equation 1;

V = (l^2)(4800-l^2)/4l = (4800l - l^3)/4

differentiating Volume V, we have;

dV/dl = (4800 - 3l^2)/4

At maximum point, dV/dl = 0

(4800 - 3l^2)/4 = 0

3l^2 = 4800

l = √(4800/3)

l = 40 cm

But

h = (4800 - l^2)/4l = (4800 - 40^2)/(4×40)

h = 3200/160

h = 20 cm

At maximum Volume;

V = (l^2)h = (40^2)×20

V = 32000 cm^3

the maximum volume of the box, V is 32000cm^3

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