When it says “largest garden” I’m assuming it means largest area.
Start with what they’ve given you: It has to be rectangular You have 56 feet of fencing
So draw a rectangle and label it. A rectangle will have 2 shorter sides and 2 longer sides. Let’s say the shorter sides are each length x and the longer sides are each length y. This rectangle is made up of fencing.
Now since you only have 56 feet of fencing that means that all the side lengths added together (ie. the perimeter) is equal to 56. Figure out the equation for the perimeter of a rectangle:
2 short sides + 2 long sides = 56 feet of fence In other words: x+x + y+y = 56 which can be rewritten: 2x + 2y = 56
You also know that you want to maximize the area of the garden, which is the area enclosed by the rectangle with side lengths x and y. The area of a rectangle is just x times y. Rewritten: A = xy
In order to maximize area, A, you need to take the derivative of A and set the derivative equal to 0 (which will give you the local max/min of your function). The problem is that you currently have 2 different variables in your function A. What you can do then, is rewrite the function A so that it only has 1 variable.
From earlier we have: 2x + 2y = 56 So rearrange it to put isolate one variable on the left side of the equals (I chose y, but you can do this with x too): 2y = 56 - 2x y = (56 - 2x)/2 y = 56/2 - 2x/2 y = 28 - x
So since we can rewrite y in terms of x, let’s substitute that into the area equation:
A = xy A = x (28 - x) A = 28x - x2
This is better because we now only have one variable. Now take the derivative of A and set it equal to 0 to find the local max:
A = 28x - x2 A’ = 28 - 2x
Set equal to 0 and solve for x: 28-2x = 0 28 = 2x 14 = x
So the short side of the rectangle, x, is 14 feet long. We also know from the work done above that y = 28-x, so we can use that to find y:
y= 28-x y= 28-14 y = 14
So the result is that the short side of the rectangle is 14 feet and the long side of the rectangle is also 14 feet.