Mathematics, 23.11.2019 18:31 shwubblz

If 1000 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

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Answer from: khaekhae10

The largest possible volume of the box is given by the maximum value of

the function for the volume of the box.

The largest possible volume of the box is approximately 3,042.9 cm³.

Reasons:

Known parameters:

Area of the box, A = 1000 cm²

Shape of the base of the box = Square

Required:

Largest possible volume of the box

Solution:

Let y represent the side length of the square base and let x represent the

height of the box, we have;

Volume, V = y²·x

Given area, A = y² + 2·y·x + 2·y·x = y² + 4·y·x

∴ 1000 = y² + 4·y·x

Which gives;

$x =\dfrac{1000 - y^2}{4 \cdot y}$

Plugging in the value of x in the volume, V formula, we get;

$V =\dfrac{y^2 \cdot \left (1000 - y^2\right)}{4 \cdot y}$

Differentiating and equating the result to zero gives;

$\dfrac{dV}{dy} = \dfrac{d}{dy} \left( \dfrac{y^2 \cdot \left (1000 - y^2\right)}{4 \cdot y} \right) =250 - \dfrac{3}{4} \cdot y^2 = 0$

$y = \sqrt{\dfrac{1000}{3} } = \dfrac{10\cdot \sqrt{30} }{3}$

Which gives;

$x = \dfrac{1000 - \left( \dfrac{10\cdot \sqrt{30} }{3} \right)^2}{4 \times \left( \dfrac{10\cdot \sqrt{30} }{3} \right)}= \dfrac{5 \cdot \sqrt{30} }{3}$

Therefore;

$V = \left( \dfrac{10\cdot \sqrt{30} }{3} \right)^2 \times \dfrac{5 \cdot \sqrt{30} }{3} = \dfrac{5000 \times \sqrt{30} }{9} \approx 3,042.9$

The largest possible volume of the box is V ≈ 3,042.9 cm³.

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Answer from: xuzixin2004

Maximum volume of the box is 3042.91 cm³

Step-by-step explanation:

Area of the material available to make a box with square base and open top = 1000 square cm.

Total surface area of the box = 2(lh + bh) + l×b

Where l = length of the box

b = width of the box

h = height of the box

If l = b, Total surface area of the box = 2(lh + lh) + l×l

Surface area = 4lh + l² = 1000

l(4h + l) = 1000

4h + l = $\frac{1000}{l}$

h = $\frac{1000}{4l}-\frac{l}{4}$

Now volume of the square box = lbh

V = l²h

V = l²( $\frac{250}{l}-\frac{l}{4}$ )

V = 250l - $\frac{l^{3}}{4}$

Now to calculate the maximum volume we have to find the derivative of the volume.

$\frac{dV}{dl}=250-\frac{3l^{2} }{4}$

By equating derivative to zero,

250 - $\frac{3l^{2}}{4}$ = 0

l = $\sqrt{\frac{250\times 4}{3}}$

l = 18.26 cm

For l = 18.26, V = 250(18.26) - $\frac{(18.26)^{3}}{4}$

V = 4565 - 1522.09

= 3042.91 cm³

Therefore, maximum volume of the box is 3042.91 cm³

Answer from: Quest

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)

Answer from: Quest

i think the answer is c sorry if it's wrong

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