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Vaselesa [24]
3 years ago
12

A manufacturer wants to make open tin boxes from pieces of tin with dimensions 8 in. by 15 in. by cutting equal squares from the

four corners and turning up the sides. Find the side of the square cutout that gives the box the largest possible volume.
0 in.
6 in.
52788.gif
52789.gif
Mathematics
1 answer:
Stells [14]3 years ago
5 0

Answer:

\frac{5}{3} in

Step-by-step explanation:

Let x be the side of square.

Length of box=8-2x

Width of box=15-2x

Height of box=x

Volume of box=L\times B\times H

Substitute the values then we get

Volume of box=V(x)=(8-2x)(15-2x)x=(15x-2x^2)(8-2x)

V(x)=120x-30x^2-16x^2+4x^3

V(x)=4x^3-46x^2+120x

Differentiate w.r.t x

V'(x)=12x^2-92x+120

V'(x)=0

12x^2-92x+120=0

3x^2-23x+30=0

3x^2-18x-5x+30=0

3x(x-6)-5(x-6)=0

(x-6)(3x-5)=0

x-6=0\implies x=6

3x-5=0\implies x=\frac{5}{3}

Again differentiate w.r.t x

V''(x)=24x-92

Substitute x=6

V''(6)=24(6)-92=52>0

Substitute x=5/3

V''(5/3)=24(5/3)-92=-52

Hence, the volume is maximum at x=\frac{5}{3}

Therefore, the side of the square ,x=\frac{5}{3} in cutout that gives the box the largest possible volume.

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Answer:

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Step-by-step explanation:

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A^c \cap (B \cap C) = ( A^c \cap B) \cap C

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