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igor_vitrenko [27]
3 years ago
8

A home improvement contractor is painting the walls and ceiling of a rectangular room. The volume of the room is 1584 cubic feet

. The cost of wall paint is $0.06 per square foot and the cost of ceiling paint is $0.11 per square foot. Find the room dimensions that will minimize the cost of the paint. (Enter your answers as a comma-separated list.) Incorrect: Your answer is incorrect. ft
Mathematics
1 answer:
liq [111]3 years ago
5 0

Answer:

The room dimensions that will minimize the cost of the paint are 12 ft x 12 ft x 11 ft.

Step-by-step explanation:

We can find first the volume equation using the formula of the volume of a box.

V= xyz

Thus we get the constraint function

1584 = xyz

Then since we are asked to minimize the cost, we can write the cost function which is the area of each one of the walls and ceiling multiplied by the painting cost.

C=0.11 xy+ 2(0.06)xz+2(0.06yz \\ C =0.11 xy+0.12xz+0.12yz

Lagrange Multipliers to find minimum cost.

We can continue finding the partial derivatives to build the system of equations required for Lagrange Multipliers method.

C_x=\lambda V_x \\ C_y = \lambda V_y \\ C_z = \lambda V_z

And the constraint function

xyz=1584

So we get

0.11y+0.12z=\lambda yz \\ 0.11x+0.12z=\lambda xz \\ 0.12x+0.12y=\lambda xy\\ xyz=1584

We can multiply each side of each equation by the dimension which is missing to get the full volume on the right side.

0.11xy+0.12xz=\lambda xyz \\ 0.11xy+0.12yz=\lambda xyz \\ 0.12xz+0.12yz=\lambda xyz

Then we can set each the equations equal to each other, so from the first one and the second equation we get

0.11xy+0.12xz= 0.11xy+0.12yz

We can subtract 0.11xy from both sides.

0.12xz=0.12yz

And we can divide both sides by 0.12z to get

x=y

We can repeat the process by setting the first and third equation equal to each other.

0.11xy+0.12xz= 0.12xz+0.12yz

We can subtract 0.12 xz from both sides

0.11xy=0.12yz

And we can solve by z

z= \cfrac{0.11x}{0.12}\\ z = \cfrac{11x}{12}

So if we replace that as well y = x on the constraint for the volume euqation we get

1584=x(x)\left(\cfrac{11}{12}x\right) \\ 1584=\cfrac{11}{12}x^3

We can then solve for x

x^3 = \cfrac{1584(12)}{11}

And taking the cube root

x = \sqrt[3]{\cfrac{1584(12)}{11}}

x = \sqrt[3]{1728}

x=12 ft

So then we can use the equations we have found for y and z in terms of x

y = x \\ y = 12 ft

And

z= \cfrac{11x}{12}\\z= \cfrac{11(12)}{12} \\ z=11ft

Then the dimensions of the room that will minimize the cost are 12ft x 12 ft x 11 ft. Since you have to enter using commas you can write 12, 12, 11, please check as well if you have to insert the units that are feet for each.

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Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

What we have to do is to isolate cos first.

\displaystyle  \large{ cos \theta =  -  \frac{1}{2} }

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>2</u>

\displaystyle \large{ \pi -  \frac{ \pi}{3}  =  \frac{3 \pi}{3}  -  \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{2 \pi}{3} }

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi  +   \frac{ \pi}{3}  =  \frac{3 \pi}{3}   +   \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{4 \pi}{3} }</u>

Both values are apart of the interval. Hence,

\displaystyle \large \boxed{ \theta =  \frac{2 \pi}{3} , \frac{4 \pi}{3} }

<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

Isolate sin(4 theta).

\displaystyle \large{sin 4 \theta =  -  \frac{1}{ \sqrt{2} } }

Rationalize the denominator.

\displaystyle \large{sin4 \theta =  -  \frac{ \sqrt{2} }{2} }

The problem here is 4 beside theta. What we are going to do is to expand the interval.

\displaystyle \large{0 \leqslant  \theta < 2 \pi}

Multiply whole by 4.

\displaystyle \large{0 \times 4 \leqslant  \theta \times 4 < 2 \pi \times 4} \\  \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi +  \frac{ \pi}{4}  =  \frac{ 4 \pi}{4}  +  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{  \frac{5 \pi}{4} }</u>

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{2 \pi -  \frac{ \pi}{4}  =  \frac{8 \pi}{4}  -  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{ \frac{7 \pi}{4} }

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

\displaystyle \large{ \theta + 2 \pi k =  \theta \:  \:  \:  \:  \:  \sf{(k  \:  \: is \:  \: integer)}}

Hence:-

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>3</u>

\displaystyle \large{ \frac{5 \pi}{4}  + 2 \pi =  \frac{13 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 4\pi =  \frac{21 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 6\pi =  \frac{29 \pi}{4} }

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{ \frac{ 7 \pi}{4}  + 2 \pi =  \frac{15 \pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 4 \pi =  \frac{23\pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 6 \pi =  \frac{31 \pi}{4} }

Therefore:-

\displaystyle \large{4 \theta =  \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4}  }

Then we divide all these values by 4.

\displaystyle \large \boxed{\theta =  \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16}  }

Let me know if you have any questions!

3 0
2 years ago
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