1) A raised bed planter with 3 sections is being designed. It will be made from

Question

1) A raised bed planter with 3 sections is being designed. It will be made from a total of 72 feet of wooden boards and use an existing retaining wall as one edge, as depicted in the following diagram. Since the back of each section of the planter will be attached to the wall, wooden boards are not needed along the back wall. ), how wide will the entire planter be, and what will be the total area of the 3-section planter? Show some work - some arithmetic. b) If each section of the planter is 15 feet long (front to back), how wide will the entire planter be, and what will be the total area of the 3-section planter? Show some work - some arithmetic. o) If each section of the planter is x feet long (front to back), how wide will the entire planter be, and what function will express the total area of the 3-section planter? is function- the constraints that insure that both dimensions d) What is the feasible domain of thi are positive?

Explanation / Answer

The perimeter of the wooden raised bed = 72 ft

Let width of each section is xft then we are given length = 6ft

Now Total perimeter of planter = 3x+6+6 = 72

=> 3x+12 = 72 => 3x = 60 => x=20 ft

Therefore the entire width = 3(20) = 60ft

And so the total area of planter = (length) (width) = (6)(60) ft2 = 360ft2

(b) Now again let width of each section = xft and length = 15ft

Then 72 = 3x + 15 + 15 = 3x+30

=> 3x = 42 => x=14 ft

Now the entire width = 3(14) = 42ft

Total area of planter = (15)(42) = 630 ft2

(c) For x ft long planter, let y be width of each section

Then 72 = 2x+ 3y => 72-2x = 3y

And so width of entire planter = 3y = 72-2x

Area of entire planter = (x)(72-2x) = 72x-2x2

x>0 And also 72x-2x2>0 => x2-36x<0

=> x(x-36)<0

=> x lies in interval (0,36) which is required feasible domain of the function.

(e) Now A(x) = 72x-2x2

A'(x) = 72-4x = 0 => x=18

Now A"(x) = -4<0

So by second derivative test, x=18 is domain that provides maximum area and Maximu area = 72(18)-2(18)2

Maximum Area = 1296-648 = 648 ft2

(f) 72x-2x2 = 300

=> 2x2-72x+300 = 0

=> x2 - 36x+150 = 0

=> x = 4.8, 31.1

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