this is section 3.8 problem 10:\na rectangle has a perimeter of 32 in. find the length and width of the…

this is section 3.8 problem 10:\na 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:\n(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=\n(b) the condition that x and y must satisfy is y=\n(c) using the condition to replace y by x in a, a can then be expressed as a function of x: a(x)=\n(d) the domain of a is , .\n(e) the only critical number of a in the domain is x=\n. we use the second-derivative test to classify the critical number as a relative maximum or minimum, or neither.\nat the critical number x=\n, the second derivative a( ) is --select-- < >. therefore at x=\n, --select--\n(f) finally, plug x=\n into the condition of x and y we obtain y=\n.\ntherefore the length and width of the rectangle under which the area is the largest are x=\n in. and y=\n in.\n

this is section 3.8 problem 10:\na 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:\n(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=\n(b) the condition that x and y must satisfy is y=\n(c) using the condition to replace y by x in a, a can then be expressed as a function of x: a(x)=\n(d) the domain of a is , .\n(e) the only critical number of a in the domain is x=\n. we use the second-derivative test to classify the critical number as a relative maximum or minimum, or neither.\nat the critical number x=\n, the second derivative a( ) is --select-- < >. therefore at x=\n, --select--\n(f) finally, plug x=\n into the condition of x and y we obtain y=\n.\ntherefore the length and width of the rectangle under which the area is the largest are x=\n in. and y=\n in.\n

Answer

Explanation:

Step1: Find the relationship between (x) and (y)

The perimeter of a rectangle (P = 2(x + y)), given (P=32), so (2(x + y)=32), then (y = 16 - x).

Step2: Express the area (A) as a function of (x)

The area of a rectangle (A=xy), substituting (y = 16 - x) into it, we get (A(x)=x(16 - x)=16x - x^{2}).

Step3: Find the domain of (A(x))

Since (x>0) and (y = 16 - x>0) (length and width are positive), solving (16 - x>0) gives (x<16). So the domain of (A) is ([0,16]).

Step4: Find the critical number

Differentiate (A(x)) with respect to (x), (A^{\prime}(x)=\frac{d}{dx}(16x - x^{2})=16 - 2x). Set (A^{\prime}(x) = 0), (16 - 2x=0), then (x = 8).

Step5: Use the second - derivative test

Differentiate (A^{\prime}(x)) to get (A^{\prime\prime}(x)=\frac{d}{dx}(16 - 2x)=- 2). Since (A^{\prime\prime}(8)=-2<0), at (x = 8), (A(x)) has a relative maximum.

Step6: Find (y)

Substitute (x = 8) into (y = 16 - x), we get (y=8).

Answer:

(a) (A = xy) (b) (y = 16 - x) (c) (A(x)=16x - x^{2}) (d) ([0,16]) (e) (x = 8), (A^{\prime\prime}(8)=-2<0) (a relative maximum) (f) (x = 8), (y = 8)