QuestionA man rows 2 miles to point $P$ and walks to house $B$ (6 miles away). Find $T(x)$ as a function of $x$ :
$T(x)=\frac{\sqrt{x^{2}-12 x+40}}{3}+\frac{x}{5}$
Is the goal to minimize $T(x)$ ?

Studdy Solution

STEP 1

Assumptions1. The man is initially miles from the nearest point $A$ on a straight shoreline. . The house is located at a point $B$ that is6 miles farther down the shoreline from point $A$ .
3. The man plans to row to a point that is between $A$ and $B$ and is $x$ miles from the house, and then walk the remainder of the distance.
4. The man can row at a rate of $3 \mathrm{mi} / \mathrm{hr}$ and can walk at a rate of $5 \mathrm{mi} / \mathrm{hr}$.
5. is the total time required to reach the house, expressed as a function of $x$ .
6. The function is given by $(x)=\frac{\sqrt{x^{}-12 x+40}}{3}+\frac{x}{5}$ .

STEP 2

The problem does not specify whether the man rows from his current location to point, and then walks from point to the house. However, based on the given function, we can infer that this is indeed the case. The function $(x)$ is composed of two terms the first term represents the time taken to row to point, and the second term represents the time taken to walk from point to the house.

STEP 3

The first term of the function $(x)$ is $\frac{\sqrt{x^{2}-12 x+40}}{3}$ . This term represents the time taken to row to point. The distance rowed is given by the expression under the square root, $x^{2}-12 x+40$ , and the speed of rowing is $3 \mathrm{mi} / \mathrm{hr}$ . Therefore, the time taken to row is the distance divided by the speed, which gives us the term $\frac{\sqrt{x^{2}-12 x+40}}{3}$ .

STEP 4

The second term of the function $(x)$ is $\frac{x}{}$ . This term represents the time taken to walk from point to the house. The distance walked is $x$ miles, and the speed of walking is $\mathrm{mi} / \mathrm{hr}$ . Therefore, the time taken to walk is the distance divided by the speed, which gives us the term $\frac{x}{}$ .

STEP 5

The function $(x)$ , therefore, represents the total time taken to reach the house by rowing to point and then walking to the house. The value of $x$ that minimizes this function will give the optimal distance from the house at which the man should stop rowing and start walking.

STEP 6

To find the value of $x$ that minimizes $(x)$ , we can take the derivative of $(x)$ with respect to $x$ and set it equal to zero. This will give us the critical points of the function, one of which will be the minimum.

STEP 7

The derivative of $(x)$ with respect to $x$ is given by $'(x)=\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5}$

STEP 8

Setting the derivative equal to zero gives us the equation $\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5}=0$

STEP 9

olving this equation for $x$ will give us the critical points of the function $(x)$ . These are the values of $x$ that could potentially minimize $(x)$ .

STEP 10

However, without a specific goal for $(x)$ , it is not clear what we are supposed to do with these critical points. If the aim is to minimize $(x)$ , then we would need to test these critical points to see which one gives the smallest value of $(x)$ . If there is another goal, then we would need more information to proceed.

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QuestionA man rows 2 miles to point $P$ and walks to house $B$ (6 miles away). Find $T(x)$ as a function of $x$ :
$T(x)=\frac{\sqrt{x^{2}-12 x+40}}{3}+\frac{x}{5}$
Is the goal to minimize $T(x)$ ?

Studdy Solution

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

The function $(x)$ , therefore, represents the total time taken to reach the house by rowing to point and then walking to the house. The value of $x$ that minimizes this function will give the optimal distance from the house at which the man should stop rowing and start walking.

STEP 6

To find the value of $x$ that minimizes $(x)$ , we can take the derivative of $(x)$ with respect to $x$ and set it equal to zero. This will give us the critical points of the function, one of which will be the minimum.

STEP 7

The derivative of $(x)$ with respect to $x$ is given by $'(x)=\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5}$

STEP 8

Setting the derivative equal to zero gives us the equation $\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5}=0$

STEP 9

olving this equation for $x$ will give us the critical points of the function $(x)$ . These are the values of $x$ that could potentially minimize $(x)$ .

STEP 10

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QuestionA man rows 2 miles to point PPP and walks to house BBB (6 miles away). Find T(x)T(x)T(x) as a function of xxx:T(x)=x2−12x+403+x5 T(x)=\frac{\sqrt{x^{2}-12 x+40}}{3}+\frac{x}{5} T(x)=3x2−12x+40​​+5x​Is the goal to minimize T(x)T(x)T(x)?

Studdy Solution

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

The function (x)(x)(x), therefore, represents the total time taken to reach the house by rowing to point and then walking to the house. The value of xxx that minimizes this function will give the optimal distance from the house at which the man should stop rowing and start walking.

STEP 6

To find the value of xxx that minimizes (x)(x)(x), we can take the derivative of (x)(x)(x) with respect to xxx and set it equal to zero. This will give us the critical points of the function, one of which will be the minimum.

STEP 7

The derivative of (x)(x)(x) with respect to xxx is given by′(x)=x−63x2−12x+40+15'(x)=\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5} ′(x)=3x2−12x+40​x−6​+51​

STEP 8

Setting the derivative equal to zero gives us the equationx−63x2−12x+40+15=0\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5}=03x2−12x+40​x−6​+51​=0

STEP 9

olving this equation for xxx will give us the critical points of the function (x)(x)(x). These are the values of xxx that could potentially minimize (x)(x)(x).

STEP 10

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QuestionA man rows 2 miles to point $P$ and walks to house $B$ (6 miles away). Find $T(x)$ as a function of $x$ :
$T(x)=\frac{\sqrt{x^{2}-12 x+40}}{3}+\frac{x}{5}$
Is the goal to minimize $T(x)$ ?

The function $(x)$ , therefore, represents the total time taken to reach the house by rowing to point and then walking to the house. The value of $x$ that minimizes this function will give the optimal distance from the house at which the man should stop rowing and start walking.

To find the value of $x$ that minimizes $(x)$ , we can take the derivative of $(x)$ with respect to $x$ and set it equal to zero. This will give us the critical points of the function, one of which will be the minimum.

The derivative of $(x)$ with respect to $x$ is given by $'(x)=\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5}$

Setting the derivative equal to zero gives us the equation $\frac{x-6}{3\sqrt{x^{2}-12 x+40}}+\frac{1}{5}=0$

olving this equation for $x$ will give us the critical points of the function $(x)$ . These are the values of $x$ that could potentially minimize $(x)$ .