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3 years ago

13

Solve the following equation on the interval [0,2 \pi) Cot(3x)=\sqrt{3}

1 answer:

Orlov [11]3 years ago

4 0

Cot(3x)=3^(1/2)

cot(x)= cos(x)/sin(x)

cot(3x)= cos(3x)/sin(3x)

look at a unit circle chart

find a value for cos that has sqrt(3) in it

which one will reduce so that the cos/sin will equal sqrt(3)?

pi/6 radians

now cot(3x) implies that x is one third this value

pi/18 is the answer

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A rectangular storage container with an open top is to have a volume of 10 m3 . then length of its base is twice the width. mate

katrin [286]

Answer:

The cost of materials for the cheapest such container is $163.54.

Step-by-step explanation:

A rectangular storage container with an open top is to have a volume of 10 m³.

The volume of the rectangle is

$\text{Volume} =\text{Length} \times \text{Width} \times \text{Height}$

Length of its base is twice the width.

Let Width be 'w'.

Length is l=2w.

Height be 'h'.

$10 =2w\times w\times h$

$10=2w^2h$

The height in terms of width is represented as,

$h=\frac{10}{2w^2}$

$h=\frac{5}{w^2}$

According to question,

The cost is 10 times the area of the base and 6 times the total area of the sides.

i.e. Cost is given by,

$C=10(L\times W)+6(2\times L\times H+2\times W\times H)$

$C=10(2w\times w)+6(2\times 2w\times \frac{5}{w^2}+2\times w\times \frac{5}{w^2})$

$C=20w^2+\frac{120}{w}+\frac{60}{w}$

$C(w)=20w^2+\frac{180}{w}$

To get the minimum value,

Differentiate the cost w.r.t 'w',

$C'(w)=20\frac{d(w^2)}{dw}+180\frac{d(w^{-1})}{dw}$

$C'(w)=20\times 2w-180 w^{-2}$

$C'(w)=40w-\frac{180}{w^2}$

To find critical points put derivate =0,

$40w-\frac{180}{w^2}=0$

$40w=\frac{180}{w^2}$

$w^3=\frac{180}{40}$

$w=\sqrt[3]{4.5}$

$w=1.65$

We find the second derivative to minimize,

$C''(w)=40\frac{d(w)}{dw}-180\frac{d(w^{-2})}{dw}$

$C''(w)=40+360(w^{-3})$

$C''(w)>0$

As $C''(w)>0$ it is the minimum cost.

The cost is minimum at w=1.65.

Substitute the values in the cost function,

$C(1.65)=20(1.65)^2+\frac{180}{1.65}$

$C(1.65)=54.45+109.09$

$C(1.65)=163.54$

Therefore, the cost of materials for the cheapest such container is $163.54.

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