Math / Calculus

QuestionSign in What's new New tab Login / Virgi About / The 10.5 Hmwk https://www.webassign.net/web/Student/Assignment-Responses/submit?dep=35488821\&tags=autosave\#question5097773_4 Mathway Mail Carlse The Coconu $A^{\prime \prime}$
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DETAILS MY NOTES TANAPMATH7 10.5.010.EP. PRACTICE ANOTHER
Consider the following closed rectangular box that has a square cross section, a capacity of $112 \mathrm{in}^{3}$ , and is constructed using the least amount of material.
Let $x$ denote the length (in inches) of the sides of the box and let $y$ denote the height (in inches) of the box. Utilize the given volume to write an equation for $y$ in terms of $x$ . $y=\square 1$
Write a function $f$ in terms of $x$ that describes the amount of material needed to create the box. $f(x)=$ $\square$ Find $f^{\prime}(x)$ and $f^{\prime \prime}(x)$ . $\begin{array}{c} f^{\prime}(x)=\square \\ f^{\prime \prime}(x)=\square \end{array}$
What are the dimensions of the box if it is constructed using the least amount of material? $\begin{array}{ll} x=\square & \text { in } \\ y=\square & \text { in } \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to find the dimensions of a square-based box with a given volume that uses the least amount of material. Watch out! Don't mix up volume and surface area!
We're minimizing surface area, not volume.

STEP 2

1. Express height in terms of side length
2. Create the surface area function
3. Find first and second derivatives
4. Minimize the surface area
5. Calculate the dimensions

STEP 3

Alright, we're given that the box has a square base with side length $x$ and height $y$ .
The volume is given as $112$ $\text{in}^3$ .
The volume of a box is length times width times height, so in our case, it's $x \cdot x \cdot y = x^2y$ .

STEP 4

Since the volume is $112$ $\text{in}^3$ , we have the equation $x^2y = 112$ .
We want to express $y$ in terms of $x$ , so we divide both sides by $x^2$ to get $y = \frac{112}{x^2}$ .
Awesome!

STEP 5

Now, let's think about the surface area.
A box has six sides.
We have two square bases, each with area $x^2$ , and four rectangular sides, each with area $xy$ .
So, the total surface area is $2x^2 + 4xy$ .

STEP 6

We already found that $y = \frac{112}{x^2}$ .
Let's substitute this into our surface area formula to get a function of just $x$ : $f(x) = 2x^2 + 4x \cdot \frac{112}{x^2} = 2x^2 + \frac{448}{x}$ .
This is the function we want to minimize!

STEP 7

Time for some calculus!
The first derivative $f'(x)$ tells us about the slope, and we'll set it to zero to find potential minimum points.
Using the power rule, we get $f'(x) = 4x - \frac{448}{x^2}$ .

STEP 8

The second derivative $f''(x)$ tells us about the concavity.
If it's positive at a critical point, that point is a minimum.
Taking the derivative of $f'(x)$ , we get $f''(x) = 4 + \frac{896}{x^3}$ .

STEP 9

We set $f'(x) = 0$ to find critical points: $4x - \frac{448}{x^2} = 0$ .
Multiplying both sides by $x^2$ gives us $4x^3 - 448 = 0$ , so $4x^3 = 448$ .
Dividing by $4$ gives $x^3 = 112$ .
Taking the cube root gives us $x = \sqrt[3]{112}$ .

STEP 10

Let's check if this is a minimum.
Plugging $x = \sqrt[3]{112}$ into $f''(x)$ , we get $f''(\sqrt[3]{112}) = 4 + \frac{896}{112} = 4 + 8 = 12$ , which is positive!
So, we have a minimum at $x = \sqrt[3]{112}$ .

STEP 11

We found $x = \sqrt[3]{112}$ inches.
Now, let's find $y$ .
We know $y = \frac{112}{x^2}$ , so $y = \frac{112}{(\sqrt[3]{112})^2} = \frac{112}{112^{2/3}} = 112^{1/3} = \sqrt[3]{112}$ inches.

STEP 12

The dimensions of the box that minimize the material used are $x = \sqrt[3]{112}$ inches and $y = \sqrt[3]{112}$ inches.

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Studdy Solution

STEP 1

What is this asking? We need to find the dimensions of a square-based box with a given volume that uses the least amount of material. Watch out! Don't mix up volume and surface area!
We're minimizing surface area, not volume.

STEP 2

1. Express height in terms of side length
2. Create the surface area function
3. Find first and second derivatives
4. Minimize the surface area
5. Calculate the dimensions

STEP 3

Alright, we're given that the box has a square base with side length $x$ and height $y$ .
The volume is given as $112$ $\text{in}^3$ .
The volume of a box is length times width times height, so in our case, it's $x \cdot x \cdot y = x^2y$ .

STEP 4

Since the volume is $112$ $\text{in}^3$ , we have the equation $x^2y = 112$ .
We want to express $y$ in terms of $x$ , so we divide both sides by $x^2$ to get $y = \frac{112}{x^2}$ .
Awesome!

STEP 5

Now, let's think about the surface area.
A box has six sides.
We have two square bases, each with area $x^2$ , and four rectangular sides, each with area $xy$ .
So, the total surface area is $2x^2 + 4xy$ .

STEP 6

We already found that $y = \frac{112}{x^2}$ .
Let's substitute this into our surface area formula to get a function of just $x$ : $f(x) = 2x^2 + 4x \cdot \frac{112}{x^2} = 2x^2 + \frac{448}{x}$ .
This is the function we want to minimize!

STEP 7

Time for some calculus!
The first derivative $f'(x)$ tells us about the slope, and we'll set it to zero to find potential minimum points.
Using the power rule, we get $f'(x) = 4x - \frac{448}{x^2}$ .

STEP 8

The second derivative $f''(x)$ tells us about the concavity.
If it's positive at a critical point, that point is a minimum.
Taking the derivative of $f'(x)$ , we get $f''(x) = 4 + \frac{896}{x^3}$ .

STEP 9

We set $f'(x) = 0$ to find critical points: $4x - \frac{448}{x^2} = 0$ .
Multiplying both sides by $x^2$ gives us $4x^3 - 448 = 0$ , so $4x^3 = 448$ .
Dividing by $4$ gives $x^3 = 112$ .
Taking the cube root gives us $x = \sqrt[3]{112}$ .

STEP 10

Let's check if this is a minimum.
Plugging $x = \sqrt[3]{112}$ into $f''(x)$ , we get $f''(\sqrt[3]{112}) = 4 + \frac{896}{112} = 4 + 8 = 12$ , which is positive!
So, we have a minimum at $x = \sqrt[3]{112}$ .

STEP 11

We found $x = \sqrt[3]{112}$ inches.
Now, let's find $y$ .
We know $y = \frac{112}{x^2}$ , so $y = \frac{112}{(\sqrt[3]{112})^2} = \frac{112}{112^{2/3}} = 112^{1/3} = \sqrt[3]{112}$ inches.

STEP 12

The dimensions of the box that minimize the material used are $x = \sqrt[3]{112}$ inches and $y = \sqrt[3]{112}$ inches.

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Studdy Solution

STEP 1

What is this asking? We need to find the dimensions of a square-based box with a given volume that uses the least amount of material. Watch out! Don't mix up volume and surface area!We're minimizing surface area, not volume.

STEP 2

1. Express height in terms of side length2. Create the surface area function3. Find first and second derivatives4. Minimize the surface area5. Calculate the dimensions

STEP 3

Alright, we're given that the box has a square base with side length xxx and height yyy.The volume is given as 112112112 in3\text{in}^3in3.The volume of a box is length times width times height, so in our case, it's x⋅x⋅y=x2yx \cdot x \cdot y = x^2yx⋅x⋅y=x2y.

STEP 4

Since the volume is 112112112 in3\text{in}^3in3, we have the equation x2y=112x^2y = 112x2y=112.We want to express yyy in terms of xxx, so we **divide both sides** by x2x^2x2 to get y=112x2y = \frac{112}{x^2}y=x2112​.Awesome!

STEP 5

Now, let's think about the surface area.A box has six sides.We have two square bases, each with area x2x^2x2, and four rectangular sides, each with area xyxyxy.So, the **total surface area** is 2x2+4xy2x^2 + 4xy2x2+4xy.

STEP 6

STEP 7

Time for some calculus!The **first derivative** f′(x)f'(x)f′(x) tells us about the slope, and we'll set it to zero to find potential minimum points.Using the power rule, we get f′(x)=4x−448x2f'(x) = 4x - \frac{448}{x^2}f′(x)=4x−x2448​.

STEP 8

The **second derivative** f′′(x)f''(x)f′′(x) tells us about the concavity.If it's positive at a critical point, that point is a minimum.Taking the derivative of f′(x)f'(x)f′(x), we get f′′(x)=4+896x3f''(x) = 4 + \frac{896}{x^3}f′′(x)=4+x3896​.

STEP 9

STEP 10

STEP 11

We found x=1123x = \sqrt[3]{112}x=3112​ inches.Now, let's find yyy.We know y=112x2y = \frac{112}{x^2}y=x2112​, so y=112(1123)2=1121122/3=1121/3=1123y = \frac{112}{(\sqrt[3]{112})^2} = \frac{112}{112^{2/3}} = 112^{1/3} = \sqrt[3]{112}y=(3112​)2112​=1122/3112​=1121/3=3112​ inches.

STEP 12

The dimensions of the box that minimize the material used are x=1123x = \sqrt[3]{112}x=3112​ inches and y=1123y = \sqrt[3]{112}y=3112​ inches.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to find the dimensions of a square-based box with a given volume that uses the least amount of material. Watch out! Don't mix up volume and surface area!
We're minimizing surface area, not volume.

1. Express height in terms of side length
2. Create the surface area function
3. Find first and second derivatives
4. Minimize the surface area
5. Calculate the dimensions

Alright, we're given that the box has a square base with side length $x$ and height $y$ .
The volume is given as $112$ $\text{in}^3$ .
The volume of a box is length times width times height, so in our case, it's $x \cdot x \cdot y = x^2y$ .

Since the volume is $112$ $\text{in}^3$ , we have the equation $x^2y = 112$ .
We want to express $y$ in terms of $x$ , so we divide both sides by $x^2$ to get $y = \frac{112}{x^2}$ .
Awesome!

Now, let's think about the surface area.
A box has six sides.
We have two square bases, each with area $x^2$ , and four rectangular sides, each with area $xy$ .
So, the total surface area is $2x^2 + 4xy$ .

Time for some calculus!
The first derivative $f'(x)$ tells us about the slope, and we'll set it to zero to find potential minimum points.
Using the power rule, we get $f'(x) = 4x - \frac{448}{x^2}$ .

The second derivative $f''(x)$ tells us about the concavity.
If it's positive at a critical point, that point is a minimum.
Taking the derivative of $f'(x)$ , we get $f''(x) = 4 + \frac{896}{x^3}$ .

We found $x = \sqrt[3]{112}$ inches.
Now, let's find $y$ .
We know $y = \frac{112}{x^2}$ , so $y = \frac{112}{(\sqrt[3]{112})^2} = \frac{112}{112^{2/3}} = 112^{1/3} = \sqrt[3]{112}$ inches.

The dimensions of the box that minimize the material used are $x = \sqrt[3]{112}$ inches and $y = \sqrt[3]{112}$ inches.