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# Let $a$ and $b$ be positive numbers. Find the length of the shortest line segment that is cut off by the first quadrant and passes through the point $(a, b)$.

## shortest length is $\sqrt{a^{2}+3 a^{4 / 3} b^{2 / 3}+3 a^{2 / 3} b^{4 / 3}+b^{2}}=\left(a^{2 / 3}+b^{2 / 3}\right)^{3 / 2}$

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Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

were given to positive numbers A and B, and we have to find the length of the shortest line segment that is cut off by the first quadrant and passes through the point a. B Just get one. Well, first of all, equation of a line that passes through the point A B with a slope of em is why minus B equals M times X minus a. Now it's intuitively mm. Has to be less than zero. Otherwise, if we had a positive slope, there would not be a line segments cut off by the first quadrant. Now set X equal to zero. We can find the coordinates of the Y intercept. So we have a Y intercept with coordinates zero mhm negative AM plus B and an X intercept. You said Y equals zero, and we have that X is equal to negative B over M plus A. This is negative. Be over em, plus a zero. Yes. Therefore, the distance from the Y intercept to the X intercept shall call D as a function of the slope M. This is the square root of the difference of the X components, which is a minus B over M squared plus B minus AM squared. Now, of course, we know that she our function d is minimized at the same time when it's square is minimized, I'll call this function F. So F is a function of em is a minus B over em squared plus B minus A m squared. Now, to find the minimum of yes will differentiate and find where this is equal to zero. So f prime of em is two times a minus B over em times derivative of the inside. This is positive. Be over m squared plus two times B minus. I am in Pampers No times the derivative of the inside, which is negative A A. Now, if you simplify, we get to over M cubed times a B M minus B squared lawyers, plus a squared them to the fourth minus a b m cubed. We set this equal to zero. We can then factor by grouping so we get to over m cubed times and then we factor b out of the 1st and 2nd terms. We have B times a M minus B and we factor and a M cubed out of the second and 33rd and fourth terms. We get a minus. Yes, B sorry. A M minus b again, Just it is equals zero. I see it's going to be factored as to over m cubed times B plus A M cubed times a M minus B equals zero. And with this factory ization, well, as you've already pointed out, Okay, One possibility is that a M minus B equals zero. This would tell us that M equals B over a, which is positive. So this is not the answer. As we already pointed out, the slope needs to be negative. Therefore, the only solution is the other factor. B plus a m cubed equals zero. And so M is equal to so mhm negative cube root of B over a. Yeah. Now, in fact, a few think about the way we factored f prime of em. It follows that f prime of them. Mhm is because I think less than zero. If em is less than negative. Cuba to be over a and f prime of them is greater than zero. If m is greater than negative Cuba, it'd be over a. Therefore it follows that f has a minimum at M equals negative cube root of B over a. Yeah, I will plug this back into our distance function, so we have f of negative cube root of B over A. This is equal to once you finally simplify it down. There's a lot of steps here. I'm skipping. Mhm, a squared plus three a to the four thirds times B to the two thirds plus three. Eight the two thirds times B to the four thirds All right plus B squared. That's so I'm skipping some steps here, my peeps. But then the length is the square root of this. Now to simplify this, that we can actually factor this as yeah a to the two thirds plus B to the two thirds cubed, just huge me boy and therefore the distance a d of negative cube root of be over a is the square root of this, which is a to the two thirds plus B to the two thirds to the three halves. Power that. And so this is our shortest length

Ohio State University

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Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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