Another pen problem A rancher is building a horse pen on the c...

Question

Another pen problem A rancher is building a horse pen on the corner of her property using 1000 ft of fencing. Because of the unusual shape of her property, the pen must be built in the shape of a trapezoid (see figure).

b. Suppose there is already a fence along the side of the property opposite the side of length y. Find the lengths of the sides that maximize the area of the pen, using 1000 ft of fencing.

Accepted Answer

A university is planning to build a new trapezoidal parking lot adjacent to an existing building using the building as one of the parking lots sites. The university has 300 ft of fencing material to maximize the parking area. What should the dimensions of X and Y be? Round the answer to two decimal places if necessary. We are given this diagram of the building. And we have 4 possible answers, being lengths in terms of X and Y. Now, to solve this, let's first look at our shape. We have the shape of a trapezoid. Now if I were to draw a diagram of the reply. We noticed that We can make this into a rectangle and a triangle. We have Y, X, Y, and we'll call this missing side of a triangle A. The sum of this length equals 300, so we have Y plus X + Y + A equals 300. Or 2 Y plus X + A equals 300. Now, also notice, Our angle is 120 degrees. And if we were to make a right angle with the right angle, We would have an angle of 30 degrees in the triangle. Now we can make a substitution for A. OK. We equals To X Tangent of 30 degrees. Now tension of 30 is a known value on the unit circle that is the square root of 3 divided by 3. Now we have a substitution for A. Let's substitute this into our equation. We have 2 Y plus X plus square root of 3 divided by 3 X equals 300. We can simplify this even further. To get X multiplied by 1, plus where the 3 divided by 3. Plus 2 Y equals 300. Now, we can make use of the area of trapezoid. This formula is given by A equals 1 F. Why plus why? Plus A Multiplied by X Which simplifies to 1/2. Multiplied by 2 Y. Plus a Multiplied by X. Now let's make a substitution. You notice from our equation near the top, we can rewrite this in terms of Y, and in fact, we can rewrite it in terms of 2 Y. 2 Y equals 300 minus X. Minus 8 Which means we have 300. Minus Parentheses X. 1 plus square the 3 divided by 3. We will now substitute this into our area equation. We have 1 F. Multiplied by 300 minus X 1 plus square the 3 divided by 3. Plus our a Which is square the 3 divided by 3 X. All multiplied by X. We can now simplify this by distributing. But we will end up getting 1/2 multiplied. By 300 minus X, multiplied by X. Which simplifies to 150 X Minus 1/2 x qui. This is our area which. Now, to find the maximum, we must find the critical points. And to find a critical point, we will set our first derivative, A equals 0. So, if A equals this equation, 150 X minus 12 X squared, A prime will be by the power rule. 150 Minus X. We said 0 equals. We get X equals. 150 Now, we can verify this by checking the second derivative, which is -1. Because this is negative one. This is a concave down parabola. This means we have a maximum, which is just what we're looking for. So no. We have X equals 150. Let's go ahead and solve for why. For Y, we can use one of our previous equations. If we look up here, You can substitute into X multiplied. By 1 + square the 3 divided by 3. Plus 2 Y equals 300. You get 150 multiplied by 1 plus where the 3 divided by 3. Plus 2 Y equals 300. We get Y equals 300 minus 150 multiplied by 1 plus the 3 divided by 3. All divided by 2. And if we simplify this, we end up getting approximately 31.70 ft. We now have our two lengths, X equals 150. Y equals 31.70. Now, if we look at our possible answers, We can see the answer to our problem. Is answer A. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Optimization

Area of a Trapezoid

Constraints in Optimization Problems

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