Math / Algebra

QuestionWord problem on combined variation 315 rapa Español
Suppose that the maximum weight that a certain type of rectangular beam can support varies inversely as its length and jointly as its width and the square of its height. Suppose also that a beam 3 inches wide, 4 inches high, and 6 feet long can support a maximum of 28 tons. What is the maximum weight that could be supported by a beam that is 5 inches wide, 4 inches high, and 20 feet long? $\square$ tons

Studdy Solution

STEP 1

What is this asking? How much weight can a 5-inch wide, 4-inch high, and 20-foot long beam hold, if a 3-inch wide, 4-inch high, and 6-foot long beam can hold 28 tons, given weight capacity depends inversely on length, and jointly on width and height squared? Watch out! Don't mix up inches and feet!
We need consistent units.
Also, remember inversely means as one thing gets bigger, the other gets smaller, and jointly means they change in the same direction.

STEP 2

1. Set up the variation equation
2. Find the constant of variation
3. Calculate the maximum weight for the new beam

STEP 3

Alright, let's define our variables!
Let $W$ be the maximum weight in tons, $L$ be the length in feet, $w$ be the width in inches, and $h$ be the height in inches.

STEP 4

The problem says the weight varies inversely with length and jointly with width and the square of height.
This translates to: $W = k \cdot \frac{w \cdot h^2}{L}$ where $k$ is our constant of variation.
This equation captures how the weight changes with the dimensions.

STEP 5

We're given that a beam with $w = 3$ inches, $h = 4$ inches, and $L = 6$ feet can support $W = 28$ tons.
Let's plug these values into our equation: $28 = k \cdot \frac{3 \cdot 4^2}{6}$

STEP 6

Now, let's solve for $k$ .
First, simplify the fraction: $28 = k \cdot \frac{3 \cdot 16}{6}$ $28 = k \cdot \frac{48}{6}$ $28 = k \cdot 8$

STEP 7

To isolate $k$ , we divide both sides by 8: $\frac{28}{8} = \frac{k \cdot 8}{8}$ $k = \frac{28}{8}$ $k = \frac{7}{2}$ So, our constant of variation is $k = \frac{7}{2}$ .

STEP 8

Now we want to find the maximum weight for a beam with $w = 5$ inches, $h = 4$ inches, and $L = 20$ feet.
Let's plug these values, along with our newly found $k = \frac{7}{2}$ , back into our variation equation: $W = \frac{7}{2} \cdot \frac{5 \cdot 4^2}{20}$

STEP 9

Time to simplify! $W = \frac{7}{2} \cdot \frac{5 \cdot 16}{20}$ $W = \frac{7}{2} \cdot \frac{80}{20}$ $W = \frac{7}{2} \cdot 4$ $W = 7 \cdot 2$ $W = 14$

STEP 10

The maximum weight the new beam can support is 14 tons.

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Studdy Solution

STEP 1

STEP 2

1. Set up the variation equation
2. Find the constant of variation
3. Calculate the maximum weight for the new beam

STEP 3

Alright, let's define our variables!
Let $W$ be the maximum weight in tons, $L$ be the length in feet, $w$ be the width in inches, and $h$ be the height in inches.

STEP 4

The problem says the weight varies inversely with length and jointly with width and the square of height.
This translates to: $W = k \cdot \frac{w \cdot h^2}{L}$ where $k$ is our constant of variation.
This equation captures how the weight changes with the dimensions.

STEP 5

We're given that a beam with $w = 3$ inches, $h = 4$ inches, and $L = 6$ feet can support $W = 28$ tons.
Let's plug these values into our equation: $28 = k \cdot \frac{3 \cdot 4^2}{6}$

STEP 6

Now, let's solve for $k$ .
First, simplify the fraction: $28 = k \cdot \frac{3 \cdot 16}{6}$ $28 = k \cdot \frac{48}{6}$ $28 = k \cdot 8$

STEP 7

To isolate $k$ , we divide both sides by 8: $\frac{28}{8} = \frac{k \cdot 8}{8}$ $k = \frac{28}{8}$ $k = \frac{7}{2}$ So, our constant of variation is $k = \frac{7}{2}$ .

STEP 8

Now we want to find the maximum weight for a beam with $w = 5$ inches, $h = 4$ inches, and $L = 20$ feet.
Let's plug these values, along with our newly found $k = \frac{7}{2}$ , back into our variation equation: $W = \frac{7}{2} \cdot \frac{5 \cdot 4^2}{20}$

STEP 9

Time to simplify! $W = \frac{7}{2} \cdot \frac{5 \cdot 16}{20}$ $W = \frac{7}{2} \cdot \frac{80}{20}$ $W = \frac{7}{2} \cdot 4$ $W = 7 \cdot 2$ $W = 14$

STEP 10

The maximum weight the new beam can support is 14 tons.

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Studdy Solution

STEP 1

STEP 2

1. Set up the variation equation2. Find the constant of variation3. Calculate the maximum weight for the new beam

STEP 3

Alright, let's **define** our variables!Let WWW be the **maximum weight** in tons, LLL be the **length** in feet, www be the **width** in inches, and hhh be the **height** in inches.

STEP 4

STEP 5

We're given that a beam with w=3w = 3w=3 inches, h=4h = 4h=4 inches, and L=6L = 6L=6 feet can support W=28W = 28W=28 tons.Let's plug these values into our equation: 28=k⋅3⋅426 28 = k \cdot \frac{3 \cdot 4^2}{6} 28=k⋅63⋅42​

STEP 6

Now, let's **solve for** kkk.First, simplify the fraction: 28=k⋅3⋅166 28 = k \cdot \frac{3 \cdot 16}{6} 28=k⋅63⋅16​ 28=k⋅486 28 = k \cdot \frac{48}{6} 28=k⋅648​28=k⋅8 28 = k \cdot 8 28=k⋅8

STEP 7

To **isolate** kkk, we divide both sides by 8: 288=k⋅88 \frac{28}{8} = \frac{k \cdot 8}{8} 828​=8k⋅8​ k=288 k = \frac{28}{8} k=828​k=72 k = \frac{7}{2} k=27​So, our **constant of variation** is k=72k = \frac{7}{2}k=27​.

STEP 8

STEP 9

Time to **simplify**! W=72⋅5⋅1620 W = \frac{7}{2} \cdot \frac{5 \cdot 16}{20} W=27​⋅205⋅16​ W=72⋅8020 W = \frac{7}{2} \cdot \frac{80}{20} W=27​⋅2080​W=72⋅4 W = \frac{7}{2} \cdot 4 W=27​⋅4W=7⋅2 W = 7 \cdot 2 W=7⋅2W=14 W = 14 W=14

STEP 10

The maximum weight the new beam can support is **14 tons**.

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1. Set up the variation equation
2. Find the constant of variation
3. Calculate the maximum weight for the new beam

Alright, let's define our variables!
Let $W$ be the maximum weight in tons, $L$ be the length in feet, $w$ be the width in inches, and $h$ be the height in inches.

We're given that a beam with $w = 3$ inches, $h = 4$ inches, and $L = 6$ feet can support $W = 28$ tons.
Let's plug these values into our equation: $28 = k \cdot \frac{3 \cdot 4^2}{6}$

Now, let's solve for $k$ .
First, simplify the fraction: $28 = k \cdot \frac{3 \cdot 16}{6}$ $28 = k \cdot \frac{48}{6}$ $28 = k \cdot 8$

To isolate $k$ , we divide both sides by 8: $\frac{28}{8} = \frac{k \cdot 8}{8}$ $k = \frac{28}{8}$ $k = \frac{7}{2}$ So, our constant of variation is $k = \frac{7}{2}$ .

Time to simplify! $W = \frac{7}{2} \cdot \frac{5 \cdot 16}{20}$ $W = \frac{7}{2} \cdot \frac{80}{20}$ $W = \frac{7}{2} \cdot 4$ $W = 7 \cdot 2$ $W = 14$

The maximum weight the new beam can support is 14 tons.