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katrin [286]
3 years ago
5

Bakery has bought 250 pounds of muffin dough. They want to make waffles or muffins in half-dozen packs out of it. Half a dozen o

f muffins requires 1 lb of dough and a pack of waffles uses
3
4
lb of dough. It take bakers 6 minutes to make a half-dozen of waffles and 3 minutes to make a half-dozen of muffins. Their profit will be $1.50 on each pack of waffles and $2.00 on each pack of muffins. How many of each should they make to maximize profit, if they have just 20 hours to do everything?
Mathematics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

250 batches of muffins and 0 waffles.

Step-by-step explanation:

-1

If we denote the number of batches of muffins as "a" and the number of batches of waffles as "b," we are then supposed to maximize the profit function

P = 2a + 1.5b

subject to the following constraints: a>=0, b>=0, a + (3/4)b <= 250, and 3a + 6b <= 1200. The third constraint can be rewritten as 4a + 3b <= 1000.

Use the simplex method on these coefficients, and you should find the maximum profit to be $500 when a = 250 and b = 0. So, make 250 batches of muffins, no waffles.

You use up all the dough, have 450 minutes left, and have $500 profit, the maximum amount.

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64

Step-by-step explanation:

hope this helps!!!

3 0
3 years ago
Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured parallel to a line
Ber [7]

Answer:

The distance is:  

\displaystyle\frac{3\sqrt{142}}{10}

Step-by-step explanation:

We re-write the equation of the line in the format:

\displaystyle\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}

Notice we divided the fraction of y by 2/2, and the fraction of z by 3/3.

In that equation, the director vector of the line is built with the denominators of the equation of the line, thus:

\displaystyle\vec{v}=\left< 3, 2, \frac{5}{3}\right>

Then the parametric equations of the line along that vector and passing through the point (-2, 3, -4) are:

x=-2+3t\\y=3+2t\\\displaystyle z=-4+\frac{5}{3}t

We plug them into the equation of the plane to get the intersection of that line and the plane, since that intersection is the image on the plane of the point (-2, 3, -4)  parallel to the given line:

\displaystyle x+y+z=3\to -2+3t+3+2t-4+\frac{5}{3}t=3

Then we solve that equation for t, to get:

\displaystyle \frac{20}{3}t-3=3\to t=\frac{9}{10}

Then plugging that value of t into the parametric equations of the line we get the coordinates of the intersection:

\displaystyle x=-2+3\left(\frac{9}{10}\right)=\frac{7}{10}\\\displaystyle y=3+2\left(\frac{9}{10}\right)=\frac{24}{5} \\\displaystyle z=-4+\frac{5}{3}\left(\frac{9}{10}\right)=-\frac{5}{2}

Then to find the distance we just use the distance formula:

\displaystyle d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

So we get:

\displaystyle d=\sqrt{\left(-2-\frac{7}{10}\right)^2+\left(3-\frac{24}{5}\right)^2+\left(-4 +\frac{5}{2}\right)^2}=\frac{3\sqrt{142}}{10}

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3 years ago
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GalinKa [24]
The correct answer is 598
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2 years ago
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What’s the answer to this math problem?
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I think it’s D
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8 0
3 years ago
Given the equation 5x − 4 = –2(3x + 2), solve for the variable. Explain each step and justify your process.
blagie [28]
A.
5x-4=-2(3x+2) \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\&#10;5x-4=-2 \times 3x-2 \times 2 \\&#10;5x-4=-6x-4 \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 6x to both sides} \\&#10;11x-4=-4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 4 to both sides} \\&#10;11x=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{divide both sides by 11} \\&#10;x=0

B.
3(2x-4)=5x-1 \\&#10;6x-12=5x-1 \\&#10;\boxed{11x-12=-1} \Leftarrow \hbox{the first mistake} \\&#10;11x=11 \\&#10;\boxed{x=11} \Leftarrow \hbox{the second mistake}

Megan's solution isn't correct.
The first mistake: she subtracted 5x from the right-hand side of the equation, but added 5x to the left-hand side.
The second mistake: she divided the right-hand side of the equation by 11, but didn't divide the left-hand side.

The correct solution:
3(2x-4)=5x-1 \ \ \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\&#10;3 \times 2x+3 \times (-4)=5x-1 \\ 6x-12=5x-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{subtract 5x from both sides} \\&#10;x-12=-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  |\hbox{add 12 to both sides} \\&#10;x=11
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3 years ago
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