Question

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.)

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 \(f(x)\):

$$
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$$

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:

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