A rectangle has a perimeter of 32 in. Find the length and width of the rectangle...

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A rectangle has a perimeter of 32 in. Find the length and width of the rectangle under which the area is the largest. Follow the steps: (a) Let the width to be x and the length to be y, then the quantity to be maximized is (expressed as a function of both x and y) A= (b) The condition that x and y must satisfy is y= (c) Using the condition to replace y by x in A, A can then be expressed as a function of x : A(x)= (d) The domain of A is [ J. (e) The only critical number of A in the domain is x= We use the Second- Derivative Test to classify the critical number as a relative maximum or minimum, or neither: At the critical number x= the second derivative A" ( ) is -Select- Therefore at x= ---Select-- (f) Finally, plug x= into the condition of x and y we obtain y= Therefore the length and width of the rectangle under which the area is the largest are x3D in. and y= in. A rectangle has a perimeter of 32 in. Find the length and width of the rectangle under which the area is the largest. Follow the steps: (a) Let the width to be x and the length to be y, then the quantity to be maximized is (expressed as a function of both x and y) A= (b) The condition that x and y must satisfy is y= (c) Using the condition to replace y by x in A, A can then be expressed as a function of x : A(x)= (d) The domain of A is [ J. (e) The only critical number of A in the domain is x= We use the Second- Derivative Test to classify the critical number as a relative maximum or minimum, or neither: At the critical number x= the second derivative A" ( ) is -Select- Therefore at x= ---Select-- (f) Finally, plug x= into the condition of x and y we obtain y= Therefore the length and width of the rectangle under which the area is the largest are x3D in. and y= in.

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