Math / Algebra

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9. [-/1 Points] DETAILS MY NOTES TANAPMATH7 10.5.017. PRACTICE ANOTHER of particle board costing $30 \mathrm{t} / \mathrm{ft}^{2}$ , what are the dimensions of the enclosure that can be constructed at a minimum cost? width $\square$ ft height $\square$ ft depth $\square$ ft Need Help? Read It Watch It Submit Answer
10. [1/1 Points] DETAILS MY NOTES TANAPMATH7 10.5.020. PREVIOUS ANSWERS PRACTICE ANOTHER

Studdy Solution

STEP 1

What is this asking? We need to find the width, height, and depth of a rectangular cabinet with a volume of 24 cubic feet, where the height is 1.5 times the width, that minimizes the cost of materials, given that the top, bottom, and sides cost $\$0.60$ per square foot and the front and back cost $\$0.30$ per square foot. Watch out! Don't mix up the different material costs and make sure to convert the height into an expression involving the width!
Also, remember to keep track of your units (feet and square feet).

STEP 2

1. Set up variables and relationships.
2. Express the cost as a function of a single variable.
3. Find the derivative of the cost function.
4. Find the critical points by setting the derivative to zero.
5. Determine the dimensions that minimize the cost.

STEP 3

Let's call the width $w$ , the height $h$ , and the depth $d$ , all measured in feet.
We're given that $h = 1.5w$ , and the volume $V$ is 24 cubic feet.
Since the volume of a rectangular prism is $V = w \cdot h \cdot d$ , we have $24 = w \cdot (1.5w) \cdot d$ , or $24 = 1.5w^2d$ .

STEP 4

The cost of the top and bottom is $2 \cdot (\$0.60) \cdot w \cdot d = \$1.20wd$ .
The cost of the sides is $2 \cdot (\$0.60) \cdot h \cdot d = \$1.20hd = \$1.20(1.5w)d = \$1.80wd$ .
The cost of the front and back is $2 \cdot (\$0.30) \cdot w \cdot h = \$0.60wh = \$0.60w(1.5w) = \$0.90w^2$ .

STEP 5

The total cost $C$ is the sum of these costs: $C = 1.20wd + 1.80wd + 0.90w^2 = 3wd + 0.9w^2$ .
From the volume equation $24 = 1.5w^2d$ , we can express $d$ in terms of $w$ : $d = \frac{24}{1.5w^2} = \frac{16}{w^2}$ .

STEP 6

Substituting this into the cost equation, we get $C(w) = 3w \cdot \frac{16}{w^2} + 0.9w^2 = \frac{48}{w} + 0.9w^2$ .

STEP 7

To minimize the cost, we need to find the critical points of $C(w)$ .
We do this by taking the derivative of $C(w)$ with respect to $w$ : $C'(w) = -\frac{48}{w^2} + 1.8w$ .

STEP 8

Setting $C'(w) = 0$ , we have $-\frac{48}{w^2} + 1.8w = 0$ .
Multiplying by $w^2$ gives $-48 + 1.8w^3 = 0$ , so $1.8w^3 = 48$ , and $w^3 = \frac{48}{1.8} = \frac{480}{18} = \frac{80}{3}$ .
Taking the cube root of both sides, we get $w = \sqrt[3]{\frac{80}{3}} \approx 2.95$ .

STEP 9

Now we can find the other dimensions. $h = 1.5w \approx 1.5 \cdot 2.95 = 4.43$ and $d = \frac{16}{w^2} \approx \frac{16}{2.95^2} \approx 1.84$ .

STEP 10

The dimensions that minimize the cost are approximately: width = 2.95 ft, height = 4.43 ft, and depth = 1.84 ft.

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Studdy Solution

STEP 1

STEP 2

1. Set up variables and relationships.
2. Express the cost as a function of a single variable.
3. Find the derivative of the cost function.
4. Find the critical points by setting the derivative to zero.
5. Determine the dimensions that minimize the cost.

STEP 3

Let's call the width $w$ , the height $h$ , and the depth $d$ , all measured in feet.
We're given that $h = 1.5w$ , and the volume $V$ is 24 cubic feet.
Since the volume of a rectangular prism is $V = w \cdot h \cdot d$ , we have $24 = w \cdot (1.5w) \cdot d$ , or $24 = 1.5w^2d$ .

STEP 4

The cost of the top and bottom is $2 \cdot (\$0.60) \cdot w \cdot d = \$1.20wd$ .
The cost of the sides is $2 \cdot (\$0.60) \cdot h \cdot d = \$1.20hd = \$1.20(1.5w)d = \$1.80wd$ .
The cost of the front and back is $2 \cdot (\$0.30) \cdot w \cdot h = \$0.60wh = \$0.60w(1.5w) = \$0.90w^2$ .

STEP 5

The total cost $C$ is the sum of these costs: $C = 1.20wd + 1.80wd + 0.90w^2 = 3wd + 0.9w^2$ .
From the volume equation $24 = 1.5w^2d$ , we can express $d$ in terms of $w$ : $d = \frac{24}{1.5w^2} = \frac{16}{w^2}$ .

STEP 6

Substituting this into the cost equation, we get $C(w) = 3w \cdot \frac{16}{w^2} + 0.9w^2 = \frac{48}{w} + 0.9w^2$ .

STEP 7

To minimize the cost, we need to find the critical points of $C(w)$ .
We do this by taking the derivative of $C(w)$ with respect to $w$ : $C'(w) = -\frac{48}{w^2} + 1.8w$ .

STEP 8

Setting $C'(w) = 0$ , we have $-\frac{48}{w^2} + 1.8w = 0$ .
Multiplying by $w^2$ gives $-48 + 1.8w^3 = 0$ , so $1.8w^3 = 48$ , and $w^3 = \frac{48}{1.8} = \frac{480}{18} = \frac{80}{3}$ .
Taking the cube root of both sides, we get $w = \sqrt[3]{\frac{80}{3}} \approx 2.95$ .

STEP 9

Now we can find the other dimensions. $h = 1.5w \approx 1.5 \cdot 2.95 = 4.43$ and $d = \frac{16}{w^2} \approx \frac{16}{2.95^2} \approx 1.84$ .

STEP 10

The dimensions that minimize the cost are approximately: width = 2.95 ft, height = 4.43 ft, and depth = 1.84 ft.

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Studdy Solution

STEP 1

STEP 2

1. Set up variables and relationships.2. Express the cost as a function of a single variable.3. Find the derivative of the cost function.4. Find the critical points by setting the derivative to zero.5. Determine the dimensions that minimize the cost.

STEP 3

STEP 4

STEP 5

STEP 6

Substituting this into the cost equation, we get C(w)=3w⋅16w2+0.9w2=48w+0.9w2C(w) = 3w \cdot \frac{16}{w^2} + 0.9w^2 = \frac{48}{w} + 0.9w^2C(w)=3w⋅w216​+0.9w2=w48​+0.9w2.

STEP 7

To **minimize** the cost, we need to find the **critical points** of C(w)C(w)C(w).We do this by taking the derivative of C(w)C(w)C(w) with respect to www: C′(w)=−48w2+1.8wC'(w) = -\frac{48}{w^2} + 1.8wC′(w)=−w248​+1.8w.

STEP 8

STEP 9

Now we can find the other dimensions. h=1.5w≈1.5⋅2.95=4.43h = 1.5w \approx 1.5 \cdot 2.95 = 4.43h=1.5w≈1.5⋅2.95=4.43 and d=16w2≈162.952≈1.84d = \frac{16}{w^2} \approx \frac{16}{2.95^2} \approx 1.84d=w216​≈2.95216​≈1.84.

STEP 10

The dimensions that minimize the cost are approximately: width = **2.95 ft**, height = **4.43 ft**, and depth = **1.84 ft**.

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1. Set up variables and relationships.
2. Express the cost as a function of a single variable.
3. Find the derivative of the cost function.
4. Find the critical points by setting the derivative to zero.
5. Determine the dimensions that minimize the cost.

Substituting this into the cost equation, we get $C(w) = 3w \cdot \frac{16}{w^2} + 0.9w^2 = \frac{48}{w} + 0.9w^2$ .

To minimize the cost, we need to find the critical points of $C(w)$ .
We do this by taking the derivative of $C(w)$ with respect to $w$ : $C'(w) = -\frac{48}{w^2} + 1.8w$ .

Now we can find the other dimensions. $h = 1.5w \approx 1.5 \cdot 2.95 = 4.43$ and $d = \frac{16}{w^2} \approx \frac{16}{2.95^2} \approx 1.84$ .

The dimensions that minimize the cost are approximately: width = 2.95 ft, height = 4.43 ft, and depth = 1.84 ft.