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2 years ago

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Emily convinced her mom to buy a giant box of her favorite cereal. Her mom doesn't think the box will fit on their shelf. The vo

lume of the box is 10{,}00010,00010, comma, 000 cm^3 3 cubed. The base of the box is 252525 cm by 101010 cm

1 answer:

Mashcka [7]2 years ago

8 0

Complete question:

Emily convinced her mom to buy a giant box of her favorite cereal. Her mom doesn't think the box will fit on their shelf. The volume of the box is 10,000 cm^3 cubed. The base of the box is 25cm by 10cm.

How tall is the box of cereal?

Answer: 40cm

Step-by-step explanation:

Given the following :

Volume = 10,000cm^3

Base of box = 25cm by 10cm,

Base of box = Length × width

Therefore:

Length = 25cm, Width = 10cm

Volume = Lengh × Width × Height

10,000cm^3 = 25cm × 10cm × Height

10,000cm^3 = 250cm^2 × Height

Height = 10,000cm^3 ÷ 250cm^2

Height = 40cm

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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:

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$Length = x$

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So:

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The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

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The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

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

Open Brackets

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

Collect Like Terms

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

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

Different A with respect to y

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

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$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

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Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

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$x = \frac{128}{y}$

$x = \frac{128}{16}$

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Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

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To double-check;

Differentiate A'

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

$A" = 1024y^{-3}$

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

The above value is:

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

This means that the calculated values are at minimum.

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