Keypad Help ARIANA RIGGS Question 12 Step of 10 points PROBL

Keypad Help ARIANA RIGGS Question: 12 Step: of (10 points) PROBLEM A rancher has 400 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parallel to one of the field\'s shorter sides. Find the dimensions that maximize the enclosed area. Write your answers as fractions reduced to lowest terms ANSWER Answer 23a 3b 3 hrs 50 mins Test Status 14 14c 14d

Solution

Let longer side be x units in length and shorter side by y units

So, total fence = x +x + y +y +y = 2x +3y = 400

Area = x*y = x(400- 2x)/3

= 400x/3 - 2x^2/3

Maximum area for the above quadratic equation occurs at vertex point given by

x = -b/2a = - (400/3 / -2*2/3)

= 100

y = ( 400 -2*100)/3 = 200/3

Dimensions are x = 100 ft ; y = 200/3 ft

Find [f(x+h) -f(x)]/h

f(x) = -5x -1

f(x+h) = -5(x+h) -1

f(x+h) - f(x) = -5(x+h) -1 - (-5x -1)

= -5x -5h -1 +5x +1 = -5h

[f(x+h) -f(x)]/h = -5h/h = -5

r(x) = 3(x-3)^1/3 -5

function is horizontallly shifted by 3 units to right

vertical streteched by 3 units

vertical shift downwards by 5 units

So, basic function we have is f(x) = (x)^1/3