A fence is to be built to enclose a rectangular area of 220 square feet. The fence along three sides is to be made of material that costs 5 dollars per foot, and the material for the fourth side costs 16 dollars per foot. Find the dimensions of the enclosure that is most economical to construct.
Dimensions: x
Dimensions: x
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First we imagine a rectangle with long side x and small side y. Obviously there are two sides of x opposite each other and similarly for y. So clearly we want to minimise the cost for y as one is $5 per foot and the other $16 per foot.
Our first equation comes from the area: x * y = 220.
Our second comes from the cost: C = 5x + 5x + 5y + 16y, where C is the cost for the whole enclosure and this is simplified as C = 10x + 21y.
Now we get rid of the x in our C cost equation using equation one which yields: x = 220 / y, so our cost equation becomes: C = 2200/y + 21y. Now we want to minimise this using calculus.
dC/dy = -2200/y^2 + 21 and now setting C = 0 to find our minimum point we get:
-2200/y^2 + 21 = 0 => 21y^2 = 2200 => y ≈ 10.24 feet. I verified that this was in fact a minimum at 10.24 as the second derivative at 10.24 is positive and hence a minimum (It's easy to check yourself by looking at the second derivative).
Now subbing this into equation one yields: x ≈ 21.48 and thus our minimum cost for the enclosure is: 10(21.48) + 21(10.24) ≈ $430.
So in brief, let the longer side of your rectangle be 21.48 feet and the smaller side 10.24 feet and you will build the most economical enclosure.
Our first equation comes from the area: x * y = 220.
Our second comes from the cost: C = 5x + 5x + 5y + 16y, where C is the cost for the whole enclosure and this is simplified as C = 10x + 21y.
Now we get rid of the x in our C cost equation using equation one which yields: x = 220 / y, so our cost equation becomes: C = 2200/y + 21y. Now we want to minimise this using calculus.
dC/dy = -2200/y^2 + 21 and now setting C = 0 to find our minimum point we get:
-2200/y^2 + 21 = 0 => 21y^2 = 2200 => y ≈ 10.24 feet. I verified that this was in fact a minimum at 10.24 as the second derivative at 10.24 is positive and hence a minimum (It's easy to check yourself by looking at the second derivative).
Now subbing this into equation one yields: x ≈ 21.48 and thus our minimum cost for the enclosure is: 10(21.48) + 21(10.24) ≈ $430.
So in brief, let the longer side of your rectangle be 21.48 feet and the smaller side 10.24 feet and you will build the most economical enclosure.