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Find an equation of the line through the point $ (3, 5) $ that cuts off the least area from the first quadrant.

$y=-\frac{5}{3} x+10$

03:32

Wen Z.

02:24

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

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not weeks, whereas to find an equation of the line through the 0.35 that cuts off the least area from the first quadrant. Very special. And that's the point. Stop. Yeah. So an equation of a line passing through the 0.35 is why minus five equals M. Times three, mine for X minus three. And I want you. And so you write this an intercept form. We have y equals M X plus five minus three m. So right, his account was joke. Joker event is as now let's find the you click on the provider. Mhm X intercept. Which very good. Well, the X intercept is the point three M minus five over m zero and the why intercept is zero five minus three m is the final. Yes, Reach change. Right now, the area in the first quadrant, this is forms of triangles. The area is one half times the base times the height and the base is the X intercept three M minus five over m. And the height is the Y intercept five minus three m. I don't know if you guys and so as a function of em, our area a So is yeah. Three M minus five. What squared over Negative. What to m so mhm. Now, in order to find from least area, we're going to take the derivative. So if a prime of em yeah, is negative three M minus five. Well, this is really a ratio. So I'm going to have to use quotient rules. We have bottom to m times the derivative, the top negative three M minus five times three. It's just minus the top, which is three m minus five squared times the derivative the bottom two all over the bottom to M squared. No, let's go give away. If we solve this equation that this is equal to zero eventually you'll find that M is equal to plus or minus five thirds and skipping a few steps here. Now, of course, yeah, If we plug in m equals positive five thirds, we actually get a negative area. This really doesn't make sense intuitively, either. If we have a positive slope, we can't have both positive and positive Y and X intercept. We need to have a negative slope for this now, substituting this back into our initial equation, we get y equals negative five thirds X plus or minus. I mean, three times negative, five thirds plus three. So the equation of the line is y equals negative. Five thirds X plus eight. Um, plus 10. Yeah. I'm sorry. This should be plus five. My mistake. This is plus five plus five, which is 10. And so this is the equation of the line through the 0.35 that cuts off the least area from the first quadrant.

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