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Find the dimensions of a rectangle with area $ 1000 m^2 $ whose perimeter is as small as possible.

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$\sqrt{1000} m$ by $\sqrt{1000} m$

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Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

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Hello. Very problem. In this way to find the dimensions of a rectangle with area and 10m. Okay who is permitted? As as small as possible If this is X. This is why so area will be X. Y. So X. Y will be equal to one 1000 So I will be equal to 1000 by X. We'll be using this to you to find the perimeter. So we will be using this to minimize the perimeter. So we have parameter equal to two and 2 x plus y which is two and 2 expressed 1000 by X. Okay so we have to minimize this. So first of all we have to get the critical point so D P by D T. So deeply by the X Will be equal to 1 -1 1000 by access quite equal to zero cannot be called to zero so one minus 1000 bikes square should be equal to zero which means one will be equal to 1000 bikes esquire by cross Multiplying Access choir will be equal to 1000 so X will be equal to plus minus 10 under oath 10 meters. So we need to check if it has maximum minimum at this point so for that we have to double differentiate B with respect to X. So if our friendship there It will be zero bless 2000 my ex cube. Okay now if we take this positive 10 to 10 it will be positive, positive which means minimum. So at this point at the point is equal to 10 to 10 it is a minimum and at the point -10 and routine it is maximum. But since actually dimension we should not take native so of an X. That is length equal to 10 Under 10 m, it will be maximum area a maximum minimum parameter. So why will be equal to 1000? So 1000 by 10 and 10. Let us multiply and divide by then. So 10 10 then and there 10 m. We have this value. Thank you.

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