Arectangular bird sanctuary is being created with one side along a straight riverbank. the remaining three sides are to be enclosed with a protective fence. if there are 8 km of fence available, find the dimension of the rectangle to maximize the area of the sanctuary

2 km wide by 4 km long with the long side parallel to the river. So we can enclose a rectangular area using just 3 sides since the 4th side is provided free by the river. I'll use the variable L to represent the length of the side that's parallel to the river and W to represent the distance from the river that parallel side is. So the total perimeter of the fence will be 2W + L. Let's write down some equations. Area enclosed A = W * L Length of L expressed in terms of W 8 = 2W + L 8 - 2W = L Now substitute (8-2W) for L in the equation for the area. A = W * L A = W * (8-2W) A = 8W - 2W^2 Since we're looking for a maximum value, that will happen when the slope of the function is equal to 0. And the first derivative of the function tells you the slope at each point. So we need to calculate the first derivative. For simple functions such as what we have, we just need to multiply the coefficient of each term by the exponent of that term, then subtract 1 from the exponent. So let's calculate the first derivative of A = 8W - 2W^2 A = 8W - 2W^2 The term 8W has an implied exponent of 1 for W, so 8 * 1 = 8. And 1 - 1 = 0. So we have 8W^0. And since W^0 = 1, we have 8 as the 1st term of the derivative. The next term is -2W^2. So -2*2 = -4 and 2-1 = 1, giving us -4W^1. And we can get rid of the explicit exponent of 1 since that's assumed to exist giving us a term of -4W. So the first derivative becomes A' = 8 - 4W Now let's solve that equation for A' = 0. A' = 8 - 4W 0 = 8 - 4W 4W = 8 W = 2 So the ideal width of the fence is 2. And looking back at the equation 8 - 2W = L, we can calculate the length as 8 - 2*2 = 8 - 4 = 4, so the length of the fence is 4. So the dimensions of the sanctuary is 4 km parallel to the river and extended 2 km from the river for a total area of 4*2 = 8 km^2. To demonstrate that is the ideal size, let's show that if the width is either smaller or larger, then we'll have a smaller sanctuary. Let's show the smaller case. For this we'll introduce a variable called "e" which represents a very tiny, but non-zero amount of length. So we'll call the width W+e or 2+e and the length will be 8 - 2*(2+e) = 8 - 4 - 2e = 4 - 2e So the area is A = WL A = (2+e)(4-2e) A = 8 -4e + 4e - 2e^2 A = 8 - 2e^2 And as you can see, as long as e is none zero the area will be smaller than 8 km^2. Now let's try a width that's less than 2 by a tiny amount. Our width is now (2-e) and that makes our length (4+2e). So A = WL A = (2-e)(4+2e) A = 8 + 4e -4e - 2e^2 A = 8 - 2e^2 And once again, if the width is slightly smaller so the length is slightly longer, the area is once again, slightly smaller. So the absolute best side for the sanctuary is 2 km wide by 4 km long.

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