Mathematics, 11.01.2020 06:31 shadowsnake

Apackaging company will only accept packages for shipment if the sum of its height and perimeter of the base is not more than 270 inches. write a function to represent the volume of a box with a square base, with sum of the height and perimeter of the base equal to 270 inches. what dimensions of such a box will give the maximum volume? what is the maximum volume?

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Answer from: CaptainCc

The dimensions that will maximize the volume and the maximum volume are;

Length = 45 inches

Length = 45 inchesWidth = 45 inches

Length = 45 inchesWidth = 45 inchesHeight = 90 inches

Length = 45 inchesWidth = 45 inchesHeight = 90 inches V_max = 182,250 in³

We are told that the sum of the height and perimeter of the base of the box is not more than 270 inches.

Thus, if the side of the base is s and the height of the box is h, then we have;

4s + h = 270

We know that formula for volume of a box is;

V = lwh

Since bottom is a square, then l = w

Thus;

V = s²h

From the perimeter equation, let's make h the subject to get;

h = 270 - 4s

Thus;

V = s²(270 - 4s)

V = 270s² - 4s³

The length of side that will maximize the box will be at dV/ds = 0.

Thus;

dV/ds = 540s - 12s²

At dV/ds = 0;

540s - 12s² = 0

12s² = 540s

s = 540/12

s = 45

Put 45 for s in h = 270 - 4s to get;

h = 270 - 4(45)

h = 270 - 180

h = 90 in

Thus, maximum volume is:

V_max = 45² × 90

V_max = 182,250 in³

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Answer from: isbella29

Volume function: V=270s^2-4s^3

Each side of the square base equal to 45 inches and height of the box of 45 inches will maximize the volume.

Maximum volume: 182,250 cubic inches.

Step-by-step explanation:

Let x represent each side of the square base and h represent height of the box.

We have been given that the with sum of the height and perimeter of the base should be equal to 270 inches. We can represent this information in an equation as:

4s+h=270 This is our constraint equation.

We know that volume of the box would be equal to area of base times height.

$V=s^2\cdot h$ This is our objective equation as we need to maximize area using the given constraint.

From constraint equation, we will get:

h=270-4s

Upon substituting this value in volume equation, we will get:

$V=s^2\cdot (270-4s)$

V=270s^2-4s^3

Therefore, our required volume function would be V=270s^2-4s^3 .

Now, we will find the first derivative of our volume function as:

$V'=\frac{d}{ds}(270s^2)-\frac{d}{ds}(4s^3)$

Let us apply power rule.

$V'=2\cdot 270s^{2-1}-4\cdot 3s^{3-1}$

$V'=540s-12s^{2}$

Let us equate our first derivative equal to zero.

$540s-12s^{2}=0$

$-12s^{2}+540s=0$

-12s(s-45)=0

$-12s=0\text{ or }(s-45)=0$

$s=0\text{ or }s=45$

Since side cannot be 0, therefore, each side of the square base would be 45 inches.

Upon substituting s=45 in equation h=270-4s , we will get:

h=270-4(45)

h=270-180

h=90

Therefore, the height of the box would be 90 inches.

The maximum volume of the box would be:

$V=(45\text{ in})^2\cdot 90\text{ in}$

$V=2025\text{ in}^2\cdot 90\text{ in}$

$V=182,250\text{ in}^3$

Therefore, the maximum volume of the box would be 182,250 cubic inches.

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