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3 years ago

A car travels 20 km southwards and then travel another 20km westward what is the displacement from the rest position

1 answer:

7 0

$\text{Hello there!}\\\\\text{We're trying to find displacement}\\\\\text{We would use the Pythagorean theorem to find the displacement:}\\a^2+b^2=c^2\\\\\text{We know that the car goes 20 km south and 20 km west}\\\\\text{When you draw it out, it should look like a triangle, you would be finding}\\\text{the length of the hypotenuse}\\\\\text{Plug in 20 to} \,\,a^2\,\, \text{and}\,\, b^2}\\\\20^2+20^2=c^2\\\\\text{Solve:}\\\\20^2+20^2=c^2\\\\400+400=c^2\\\\800=c^2\\\\\sqrt800=\sqrt c\\$

$\large\boxed{28.28}\\\\\normalsize\text{The displacment would be 28.28}$

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Answer:

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As per the question:

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Let the height of the bin be 'h'

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Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

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Therefore, your correct answer would be "all of the above".

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