**Help with a calculus problem – how to set this up and figure?**

*A rectangular field is fenced along the bank of a river, and no fence is required* along the River. The fence material is $ 8 per foot run at both ends and $ 12 per foot for the side running parallel to the river, $ 3,600 worth around will use. Find the dimensions of the field of maximum area that can be closed. What is the greatest possible area?

That term is and (parallel to the river) and two widths to be x and x + y = 2x cost perimeter = 2 (8x) + 12y = 3600 16x = 12 and 3600 3600-12y = 16x = 300-4x / 3 x Area = Area = X [300-4x / 3] A = 300x-4x ^ 2.3 Differentiate wrt x dA / dx = 300 – 8x / 3 = 0 8x / 3 = 300 8x = 900 x = 900 / 8 = 112.5 and = 300 -4 (112.5) / 3 = 150 The dimensions that maximizes the area of 150 meters parallel to the river and 112.5 feet for the two extremes. 2A/dx d ^ ^ 2 =- 8.3 <0 This verifies that the area has been maximized.

**Derivatives of Logs and e^ Problem Set 1 – Calculus**