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Alex Ar [27]
3 years ago
8

A box is to be constructed from a sheet of cardboard that is 20 cm by 50 cm by cutting out squares of length x by x from each co

rner and bending up the sides. What is the maximum volume this box could have? (Round your answer to two decimal places. Do not include units, for example, 10.22 cm would be 10.22.
Mathematics
1 answer:
Nezavi [6.7K]3 years ago
4 0

We are given the dimensions of the box:

l = 50 cm

w = 20 cm

 

We know that the volume of a box is:

V = l w h

where h = x

 

However since x cuts is made on both all sides of the box, therefore the new dimensions would be:

V = (l – 2x) (w – 2x) x

V = (50 – 2x) (20 – 2x) x

V = 1000x – 100x^2 – 40x^2 + 4x^3

V = 4x^3 – 140x^2 + 1000x

 

<span>To get the maxima value, we get the 1st derivative of the function then set dV/dx = 0 to solve for x:</span>

dV / dx = 12x^2 – 280x + 1000

12x^2 – 280x + 1000 = 0

Transpose 1000 to the right side and divide everything by 12:

x^2 – (280/12)x = -(1000/12)

Completing the square:

x^2 – (280/12)x + (78400/576) = -(1000/12) + (78400/576)

[x – (280/24)]^2 = 52.78

x – (280/24) = ±7.26

<span>x =  (280/24) ± 7.26</span>

x = 4.40, 18.93

x cannot be 18.93 since this would result in a negative value of 20 – 2x, therefore:

x = 4.40 cm

 

Calculating for the volume:

V = (50 – 2*4.4) (20 – 2*4.4) (4.4)

<span>V = 2030.34 cm^3</span>

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A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its
VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate (\dot V), measured in liters per hour, is directly proportional to draining time (t), measured in hours. That is:

\dot V \propto \frac{1}{t}

\dot V = \frac{k}{t} (Eq. 1)

Where k is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) <em>Large and small drains are opened</em>

\dot V_{s}+\dot V_{l} = \frac{k}{2} (Eq. 2)

\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}

(ii) <em>Only the small drain is opened</em>

\dot V_{s} = \frac{k}{t_{l}+3} (Eq. 3)

\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}

(iii) <em>Only the big drain is opened</em>

\dot V_{l} = \frac{k}{t_{l}} (Eq. 4)

\frac{\dot V_{l}}{k}  = \frac{1}{t_{l}}

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}

\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}

2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}

t_{l}^{2}-t_{l}-3 = 0 (Eq. 5)

Whose roots are determined by the Quadratic Formula:

t_{l,1}\approx 2.303\,h and t_{l,2} \approx -1.302\,h

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

t_{s} = 2.303\,h+3\,h

t_{s} = 5.303\,h

The draining time when only the small drain is opened is 5.303 hours.

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3 years ago
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