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# Another pen problem A rancher is building a horse pen on ISBN: 9780321570567 2

## Solution for problem 50E Chapter 4.4

Calculus: Early Transcendentals | 1st Edition

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Problem 50E

Another pen problem A rancher is building a horse pen on the corner of her property using 1000 ft of fencing. Because of the unusual shape of her property, the pen must be built in the shape of a trapezoid (see figure). a. Determine the lengths of the sides that maximize the area of the pen. b. Suppose there is already a fence along the side of the property opposite the side of ? length y? . Find the lengths of the sides that maximize the area of the pen, using 1000 ft of fencing.

Step-by-Step Solution:

Solution Step 1 The given diagram is Step 2: (a) otal length of the pen l = 1000ft Here the value of Z,sis z = y tan 30 y s = 0 sin 30 We have to find the lengths of the sides that maximizes the area of the pen Total length of the trapezoid, 1000 = 2x+y+z+s = 2x+y+ y + y tan 30 sin 30 = 2x+y+ 3y+2y = 2x+3y+ 3y = 2x+ 3y( 3+1) 3 x = 500 2 y( 3+1) Step 3 Area of the trapezoid is the sum of area of triangle and rectangle,therefore 1 A = xy+ z2 = xy+ ( 3y)y 2 1 2 = xy+ 2 3y Which is objective function Substituting for y the objective function becomes A = xy+ z2 3 1 2 = (500 2y( 3+1))y+ 2 3y = (500 3y( 3+ ))y 2 2 Step 4 A is the function of single variable y To maximize A we need to differentiate it with respect to y and equate it to zero A’=0 3 1 (5002 ( 2+ ))y =20 3( 3+ ))y = 500 2 1000 y = 6+ 3 Step 5 So x = 500 3+3 1000 2 6+ 3 3+3 = 500(1 6+ 3) 6+ 3 33 = 500( 6+ 3 = 1500 6+ 3

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##### ISBN: 9780321570567

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