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Oksana_A [137]
3 years ago
11

a container with a square bottom, rectangular sides and no top is to e constructed to have a volume of 5 m^3. material for the b

ottom costs $10 per square meter and material for the sides costs $8 per square meter. find the dimensions of the least expensive container
Mathematics
2 answers:
VLD [36.1K]3 years ago
7 0

Answer:

2m\times 2m\times \frac{5}{4}m

Step-by-step explanation:

We are given that a container with square bottom.

Let side of square=x

Height of container=h

Volume of container=5m^3

Cost of 1 square meter of material for the bottom=$10

Cost of 1 square meter of material for side=$8

We have to find the dimension of least expensive container.

Volume of container=x^2h

x^2h=5

h=\frac{5}{x^2}

Surface area of container=x^2+2(x+x)h=x^2+4xh

Cost=C(x)=10x^2+8(4xh)=10x^2+32x(\frac{5}{x^2})=10x^2+\frac{160}{x}

C(x)=10x^2+\frac{160}{x}

Differentiate w.r.t x

\frac{dC}{dx}=20x-\frac{160}{x^2}

\frac{dC}{dx}=0

20x-\frac{160}{x^2}=0

\frac{160}{x^2}=20x

x^3=\frac{160}{20}=8

x=\sqrt[3]{8}=2

Because side of container is always positive.

Again differentiate w.r.t x

\frac{d^2C}{dx^2}=20+\frac{320}{x^3}

Substitute x=2

\frac{d^2C}{dx^2}=20+\frac{320}{2^3}=60>0

Hence, the cost is minimum at x=2

Substitute the value of x

h=\frac{5}{2^2}=\frac{5}{4}m

Hence, the dimensions of the least expensive container are

2m\times 2m\times \frac{5}{4}m

MaRussiya [10]3 years ago
4 0

Answer:

1.26 m , 3.15 m

Step-by-step explanation:

Let the side of the square base is y.

Height of the container is h.

Area of base = y x y = y²

Area of side walls = 4 x y x h = 4yh

Volume of the container, V = y²h

According to the question, the volume of the container is 5 m³

So, 5 = y²h

h = 5 / y²   .... (1 )

Cost of bottom, C1 = 10 y²

Cost of side walls, C2 = 8 yh

Total cost, C = 10 y² + 8yh

Substitute the value of h from equation (1)

C = 10 y² + 8 y x 5 / y²

C = 10y² + 40 / y

Differentiate it with respect to y

dC/dy = 20 y - 40/y²

For maxima and minima, dC/dy = 0

20 y = 40/y²

y³ = 2

y = 1.26 m

So, h = 5 / (1.26 x 1.26) = 3.15

Thus, the side of base is 1.26 m and height is 3.15 m to minimize the cost.

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The parabolic motion is an illustration of a quadratic function

The equation that models that path of the rocket is y = -16/31x^2 + 256/31x - 880/31

<h3>How to model the function?</h3>

Given that:

x stands for time and y stands for height in feet

So, we have the following coordinate points

(x,y) = (5,0), (11,0) and (10,80)

A parabolic motion is represented as:

y =ax^2 + bx + c

At (5,0), we have:

25a + 5b + c = 0

c= -25a - 5b

At (11,0), we have:

121a + 11b + c = 0

Substitute c= -25a - 5b

121a + 11b -25a - 5b = 0

Simpify

96a + 6b = 0

At (10,80), we have:

100a + 10b + c = 80

Substitute c= -25a - 5b

100a + 10b - 25a -5b = 80

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Make b the subject

b = 15a + 16

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b = 15a + 16

So, we have:

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[tex]b =256/31

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This gives

c= 25*16/31 - 5 * 256/31

Take LCM

c= -880/31

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y =ax^2 + bx + c

This gives

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Hence, the equation that models that path of the rocket is y = -16/31x^2 + 256/31x - 880/31

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PilotLPTM [1.2K]

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