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3 years ago

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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

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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