The owner of a used tire store wants to construct a fence to enclose a rectangular outdoor storage area adjacent to the store, using part of the side of the store (which is 220 feet long) for part of one of the sides. (See the figure below.) There are 430 feet of fencing available to complete the job. Find the length of the sides parallel to the store and perpendicular that will maximize the total area of the outdoor enclosure.
Length of parallel side(s)=?
Length of perpendicular sides=?
let the side parallel to the store by y
let the two other sides be x
y + 2x = 430
y = 430 - 2x
area = xy
= x(430 - 2x)
= 430x - 2x^2
d(area)/dx = 430 - 4x = 0 for a max area
4x = 430
x = 107.5
I will let you finish it.
333
The owner of a lumber store wants to construct a fence to enclose a rectangular outdoor storage area adjacent to the store, using part of the side of the store (which is 220 feet long) for part of one of the sides. (See the figure below.) There are 430 feet of fencing available to complete the job. Find the length of the sides parallel to the store and perpendicular that will maximize the total area of the outdoor enclosure.
Well, before I give you an answer, let me just say that the owner of the used tire store is really trying to tread carefully with this fence idea. But don't worry, I'll try to steer you in the right direction.
Let's call the length of the sides parallel to the store "x" and the length of the perpendicular sides "y." Now, we know that one side of the rectangular enclosure will be 220 feet (since it's part of the side of the store), and we have 430 feet of fencing in total.
So, the equation we can set up is: x + 2y = 430 - 220
Now, let's focus on the area. The area of a rectangle is simply the length multiplied by the width. In this case, the width is x, and the length is y. So, the total area is A = xy.
To maximize the area, we can use the fact that the area of a rectangle is maximized when its sides are equal. So, we can set x = y.
Now, let's substitute this value of x into our first equation: 2y + y = 430 - 220
Simplifying, we get: 3y = 210
Dividing both sides by 3, we find that y = 70 feet.
Since x = y, the length of the sides parallel to the store is also 70 feet.
So, the length of the parallel side(s) is 70 feet, and the length of the perpendicular sides is also 70 feet.
Now we can all agree that this fence will certainly make a good enclosure for those used tires!
To find the length of the sides parallel to the store and perpendicular, we can apply the concept of optimization. The total area of the outdoor enclosure will be maximized when the dimensions of the enclosure result in the largest possible area.
Let's denote the length of the sides parallel to the store as x and the length of the perpendicular sides as y. Since one side of the enclosure will be part of the 220-foot-long side of the store, the other side parallel to the store will also have a length of x.
Now, let's set up the equation based on the given information and constraints. The sum of the lengths of the sides requires a total of 430 feet of fencing:
2x + y + 220 = 430
Simplifying this equation, we get:
2x + y = 210
We want to maximize the area, which is given by the formula:
Area = x * y
Now, we can express the area in terms of a single variable by rearranging the equation above. From the constraint equation, we have y = 210 - 2x, so substituting this into the area equation:
Area = x * (210 - 2x)
To find the maximum area, we need to find the critical points of this quadratic function. Taking the derivative of the area equation with respect to x, we get:
d(Area)/dx = 210 - 4x
Setting this derivative equal to zero and solving for x, we have:
210 - 4x = 0
4x = 210
x = 210/4
x = 52.5
Now, we can substitute this value of x back into the constraint equation to find y:
2(52.5) + y = 210
105 + y = 210
y = 210 - 105
y = 105
Therefore, the length of the sides parallel to the store is 52.5 feet, and the length of the perpendicular sides is 105 feet.