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# Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side $L$ if one side of the rectangle lies on the base of the triangle.

## Maximum area $=\frac{\sqrt{3} L^{2}}{8}$Length $=\frac{L}{2}$Width $=\frac{\sqrt{3} L}{4}$

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So we have a question and this went to find the dimensions of the rectangle of largest area. Oh okay. That could be described in an equator triangle. This is it. This is an equilateral triangle. Okay. Yeah. Mhm. Okay. So if abc is any critical triangle and this is the rectangle, suppose X. And why rectangle? Okay. Uh If one side of England base, so we have to find the damages of the tangle of largest area. So area of the rectangle is x. Com away area should be the largest. And this is and this is also if this is l this is X. So this will pay this part will be l minus X by two And this part will be held -X by two. Okay, because the whole thing some Okay, yes, no, this is 60 degrees 60°. So if we apply the trigonometry, Okay, trigonometry over here in the strangle this is why. So we can say that 10 60° equal to Vibe, I l minus X by 2. 10 sixties. Route three Equal to two I buy l minus x. Yeah. Ellis constant. So we have to our motivates to convert this in a single variable from here, we can do that. So why will be called to Route three x 2, L minus X. So you just plug in here a call to X into Y accent too. That route three x 2. Route three x 2 and L -X from here. So this is a route three x 2, L x minus access squad. We have to maximize the minutemen for maximum minimum. We have to differentiate this with respect to X. And we'll be finding first the critical point which will be we can find by equating the by T X equal to zero. So X equal to alibi too. No, it is just be sure that X equals 12 x two is a point of maximum or minimum. So, let us double different shape this again differentiating the rdX the Squire A by the X squared. So this will be route three x 2 Into -2. So -2 or three which is negative. So maximum X equal to alibi too. And why call to route three x 2 L -X. All right. Call to Rule three x 2. L- L x two. So Rule three way too. Al by two. So dimensions are length X equal to Al by two. And we're Why call to Ruth Real by four. Yeah. Okay. So these are the dimensions. Thank you.

Chandigarh University

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