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Sujay
You are a farmer, and you are trying to make the most efficient use of your resources. You have invested enough capital for a barn house which is 50 feet by 70 feet (its base) and 100 feet of fencing for your animals. You decide to have an outdoor lot for your animals. Your goal is to have the lot with the most area with the least amount of fencing (optimize). The lot must be rectangular, and one side of the rectangular may be a wall from the barn. A picture will be posted below.
Is it 40 feet by 70 feet, or 50 feet by 70 feet? Your diagram contradicts your question.
50 feet, sorry about that
so you will definitely want to use the 70 foot wall of the barn as part of your fence instead of the 50 foot wall. this brings your total fencing material to 170 feet (100 feet of fencing plus 70 feet from the barn). Then, you find the dimensions for the length and width of the fenced area in terms of x. Use the formula for perimeter to find these. P=170=2L+2W Let L=x, Then W=85-x Then plug into the equation for area. A=LW=x(85-x)=85x-x^2 Take the derivative of the formula for area which is 85-2x. Find your critical point(s) which turns out to be 85/2. Then plug in the values of x at the critical point and at the x- and y- intercepts of the area equation. The highest value will be your answer.
ralph94, there is a contradiction in your solution. If you have taken one length of the rectangle to be 70 ft, the opposite side is also 70 ft, leaving us with only 30 ft of fencing, which makes the other two sides 15 ft. So your dimensions will be 70 X 15. I am not sure if the following solution is correct (I really dislike max/min) Let x = the side opposite the barn wall. Keeping in mind the restriction: \[x \le 70\] So let the adjacent side = y. Then we get that the perimeter which is 100 ft: \[P = x + 2y = 100\] Area we know will be \[A = x \times y \] From the perimeter equation we get: \[x = 100 - 2y\] Plugging this into the Area equation we get: \[A = y \times (100 - 2y)\] Taking the derivative we get: \[A ^{\prime} = 100 - 4y \] Setting this equal to 0, we get y = 25, and x = 50. Maximum area will be 50 x 25 which is 1250.
Calculus is great, but why not use an easier method? The Area of the enclosure is a rectangle, so A = l x w. If the length is l, the w must be .5(100 - l). Therefore the Area = l(.5(100 - l)) = .5l(100 - l) = -.5l^2 + 50l. This graphs to a parabola, the vertex being at (50, 1250) which represents the maximum area. Therefore when the length is 50 (and the width must then be 25 for the enclosure), the Area will be the maximum at 1250. This is a good old Algebra 2 problem!