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yaroslaw [1]
2 years ago
15

At the frozen yogurt shop, a machine fills cups with 4 ounces of frozen yogurt before adding the toppings. After the cups are fi

lled, customers add as many toppings as they want without exceeding a total weight of 6 ounces. Which of the following represent the acceptable weights for the frozen yogurt options? Select all that apply.
Mathematics
2 answers:
qwelly [4]2 years ago
8 0
Is this for algebra 1 if so I believe it's 4,5,6
timofeeve [1]2 years ago
7 0
In this question, every cups will be filled with 4 ounces yogurt. That mean, the lowest possible of the cups weight would be 4 ounces. After that the customer can the topping without exceeding 6 ounces of total weight. Since it total weight, that means from the 6 ounces there should be 4 ounces of yogurt. Then the maximum weight is 6 ounces
Minimum weight is 4 ounces and maximum weight is 6 ounces, so the answer would be 4,5,6 or any number between 4-6
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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

Area = 128

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Length = x

Width = y

So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

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A = (\frac{128}{y} + 2) * (y + 4)

Open Brackets

A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

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To calculate the smallest possible value of y, we have to apply calculus.

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A' = -512y^{-2} + 2

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A' = 0

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Multiply through by y^2

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512 = 2y^2

Divide through by 2

256=y^2

Take square roots of both sides

\sqrt{256=y^2

16=y

y = 16

Recall that:

x = \frac{128}{y}

x = \frac{128}{16}

x = 8

Recall that the new dimensions are:

Length = x + 2

Width = y + 4

So:

Length = 8 + 2

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Width = 16 + 4

Width = 20

To double-check;

Differentiate A'

A' = -512y^{-2} + 2

A" = -2 * -512y^{-3}

A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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Answer:

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