find the point on the line $\\frac{x}{5}+\\frac{y}{10}=1$ that is closest to $(3,0)$.the point on the line…

find the point on the line $\\frac{x}{5}+\\frac{y}{10}=1$ that is closest to $(3,0)$.the point on the line $\\frac{x}{5}+\\frac{y}{10}=1$ that is closest to $(3,0)$ is(type an ordered pair, using integers or fractions.)

find the point on the line $\\frac{x}{5}+\\frac{y}{10}=1$ that is closest to $(3,0)$.the point on the line $\\frac{x}{5}+\\frac{y}{10}=1$ that is closest to $(3,0)$ is(type an ordered pair, using integers or fractions.)

Answer

Explanation:

Step1: Rewrite the line equation

The line equation (\frac{x}{5}+\frac{y}{10}=1) can be rewritten as (y = 10 - 2x).

Step2: Set up the distance formula

The distance (d) between a point ((x,y)) on the line and the point ((3,0)) is given by (d=\sqrt{(x - 3)^2+(y - 0)^2}). Substitute (y=10 - 2x) into the distance formula: (d=\sqrt{(x - 3)^2+(10 - 2x)^2}). To simplify calculations, we can minimize the square of the distance function (f(x)=(x - 3)^2+(10 - 2x)^2).

Step3: Expand the function

Expand (f(x)): [ \begin{align*} f(x)&=(x^{2}-6x + 9)+(100-40x + 4x^{2})\ &=x^{2}-6x + 9+100-40x + 4x^{2}\ &=5x^{2}-46x + 109 \end{align*} ]

Step4: Find the derivative

Differentiate (f(x)) with respect to (x). Using the power rule ((x^{n})^\prime=nx^{n - 1}), (f^\prime(x)=10x-46).

Step5: Set the derivative equal to zero

Set (f^\prime(x) = 0), so (10x-46=0). Solve for (x): (10x=46), (x=\frac{23}{5}).

Step6: Find the corresponding (y) - value

Substitute (x = \frac{23}{5}) into (y = 10-2x). (y=10-2\times\frac{23}{5}=10-\frac{46}{5}=\frac{50 - 46}{5}=\frac{4}{5}).

Answer:

((\frac{23}{5},\frac{4}{5}))