Renna M. answered • 05/05/20
Calculus Tutor with 9+ Years Experience Teaching High School and Colle
The first step in this problem is to gather all the information and decide if this problem can be solved.
600 u2 of land to be enclosed by fence
Area inside plot should be maximized, amount of fence should be minimized
This type of problem is very common in differential Calculus and is usually found in textbooks under the section "Applications of Differential Calculus - Optimization". You might have some experience with these types of problems from Algebra 2 where quadratic equations were used for optimization.
If we draw a quick sketch, we get something like this:
_____ _____
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The two rectangles could be different sizes but if one rectangle size maximizes area then it would make sense to use that some rectangle for both sides.
Let's put some variables into our picture so we can define our constraints.
y y
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x | x | x |
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Now we just need an equation to connect our unknowns (how much fence to use) to our constraint (it has to have an exact area of 600 u2.
A = x*2y = area = 600
P = 3x + 4y = perimeter = a minimum
If you notice above, P is an equation with two inputs (x and y). While that wouldn't be a problem in Multivariable Calculus, right now you probably only know how to do equations with one variable. That means we need the perimeter equation to have just an x or a y. Thankfully we have another equation with x and y which means substitution might help.
First we solve the area equation for y:
y = 600/2x = 300/x
and then we substitute that result into our perimeter equation:
P = 3x + 4 (300/x)
P = 3x + 1200/x
It might not look like much but it turns out that the perimeter is actually related to just one variable x. If x changes, then our perimeter will also change (because we set our problem up to always have an area of exactly 600).
*Note: we could also have an equation for perimeter where the input variable is y. However, most students feel more comfortable with x being the input so we will use that for this problem. If you want extra practice trying finding P(y).
Our perimeter equation is: P(x) = 3x + 1200/x OR P(x) = 3x +1200x-1
We can find what value of x makes our perimeter as small as possible (a minimum) by graphing or by using Calculus.
In Calculus we know that maximums and minimums always occur when the derivative is zero. This is because the slope (rate of change) at these points in a graph are always flat or zero. Our derivative is
P'(x) = 3 - 1200x-2
When set equal to zero, we can solve for x:
3-1200x-2 = 0
1-400x-2 = 0
1 = 400/x2
x2 = 400
x = ± 20
Our fence length can't be a negative number so our answer must be 20 u.
If our x length is 20, then our y length must be 300/x to ensure our area is 600 u2.
The y length is 300/(20) = 15
Let's put all of our numbers into our picture to see if it make sense.
15 15
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20 | 20 | 20 |
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15 15
Sure enough our area is 600.
To be absolutely confident in our answer, we can try a few different values and see how our Area and Perimeter values change.
Here's what we have right now:
A(20, 15) = 600
P(20, 15) = 120
How about we try something a little bigger like 20.5 for x and 14.63 for y?
A(20.5, 14.63) = 600
P(20.5, 14.63) = 120.02
How about we try something a little smaller like 19.5 for x and 15.38 for y?
A(19.5, 15.38) = 600
P(19.5, 15.38) = 120.02
It looks like if we try to change x, even a little, to make it bigger or smaller the perimeter only gets bigger.
You might have guessed at the beginning that a square is the best shape to use to maximize area, so let's try that as well.
Each rectangle is 300 u2. We want a square so each side would be √(300) = 17.32
17.32 17.32
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17.32 | 17.32 | 17.32 |
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17.32 17.32
A(17.32, 17.32) = 600
P(17.32, 17.32) = 121.24
It looks like a square is pretty close but our above answer is still the best.
