My friend has $80 to spend on a fence for her rectangular garden. She wants to use cedar fencing which costs $8/foot on one side, and cheaper metal fencing which costs $2/foot for the other three sides. What are the dimensions of the garden with the largest area she can enclose? What is the largest area that can be enclosed?

I have the perimeter as 10x+4y=80. From this I get y=20-(5x/2). So I plug that into the formula WidthxLength for Area and get the equation x(20-5x/2), which gives me 8 and 0 as the values for x, of which 8 is the only feasible. But when plugged back in to the equation for y, I get 0 which is obviously impossible. What a I missing?

Your algebra is correct. Your logic is wrong, however.

x=0,8 are roots of the function. That is, where the area is zero.

If you rewrite it as

20x - 5/2 x^2

you see it is a parabola with its vertex at x = -b/2a = 20/5 = 4
So, y=10

and the largest area is 4*10 = 40

http://www.wolframalpha.com/input/?i=x%2820-5x%2F2%29

To find the dimensions of the garden with the largest area, we need to optimize the equation for area, given the constraint of the amount of money available for the fence.

Let's start by breaking down the problem and formulating the equations:

Let x be the length of the garden (which will be fenced using the more expensive cedar fencing), and let y be the width (which will be fenced using the cheaper metal fencing).

The cost for the cedar fencing would be 8 times the length of the garden: Cost of cedar fencing = 8x
The cost for the metal fencing would be 2 times the perimeter of the garden, minus the cost of the cedar fencing: Cost of metal fencing = 2(2x + y) - 8x

We know that the total cost of the fence should not exceed $80, so we can set up the following equation:

8x + 2(2x + y) - 8x = 80

Simplifying the equation gives us:
4x + 2y = 80

Now we can rewrite this equation in terms of y:
2y = 80 - 4x
y = 40 - 2x

To find the largest area, we need to maximize the expression for area, which is given by length multiplied by width:
Area = x * y

Substituting the value of y from the equation above, we get:
Area = x * (40 - 2x)
Area = 40x - 2x^2

To find the dimensions that maximize the area, we need to find the critical points of the area equation. We can do this by taking the derivative of the area equation with respect to x and setting it equal to zero:

d(Area)/dx = 40 - 4x = 0

Solving this equation, we find that x = 10. This means that when x = 10, the value of y can be calculated as y = 40 - 2x = 40 - 20 = 20.

Therefore, the dimensions of the garden with the largest area are a length of 10 feet and a width of 20 feet. To find the largest area, we can substitute these values back into the area equation:

Area = 10 * 20 = 200 square feet.

So the largest area that can be enclosed is 200 square feet, and the dimensions of the garden are 10 feet by 20 feet.