Question

this is section 3.8 problem 10:
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 , .
(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 x=
in. and y=
in.

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) \(x = 8\), \(y = 8\)