Confused on a complicated math problem...?

Question

Im really confused on this math problem, and i need some help. here gos: The points of th graphs of xy=20 and x^2+ y^2=41 are joined to form a rectangle. Find the area of the rectangle. Include all points of intersection and express all answer exactly(no decimals).

Answer 1

There is a direct way to find the values for x and y which make this true. Just express x or y in terms of the other variable from the equation xy = 20 and then substitute it into the second equation. After doing some manipulating, you can solve for the values for those two variables.

y = 20/x
x² + y² = 41 ----> x² + (20/x)² = 41
x² = 41 - (20/x)²
x² = 41 - 400/x²
x² * x² = 41x² - 400
x^4 - 41x² + 400 = 0

The last equation is factorable into this:

(x² - 16)(x² - 25) = 0.

We can factor each of the above factors even further:

(x² - 16) = (x + 4)(x - 4)
Setting each one of these equal to 0, we get x = ±4.

(x² - 25) = (x + 5)(x - 5)
Setting each one of these equal to 0, we get x = ±5.

Since the product of x and y is a positive number, then we must use either both negative or both positive numbers when plotting the graph of the equation xy = 20. That means the solutions lie only in the first and third quadrant of the Cartesian plane.

Now we need to find the area of the rectangle. Note that the slopes of the two lines connecting the two sets of points (4, 5) and (5, 4) and (-5, -4) and (-4, -5) are both -1. So the length of the line segment joing either set of points is d = √[(1)² + (-1)²] = √2 units long. This line segment joins the points which are common to both of your equations, and forms the width, W, of the rectangle whose area you seek. To see why, plot the points on graph paper or a carefully constructed Cartesian system on note paper. You will see that the lines connecting the upper and lower points are parallel to each other. Furthermore, the two lines are also perpendicular to the lines containing those points. Therefore, the four lines form a rectangle. From this we conclude W = √2.

Now we have to find the length of the rectangle. To do that we first evenly divide the line segment whose length we just found by using a perpendicular line segment which runs to the origin, (0, 0), which also happens to be the center of the circle. The origin and the points of intersection of the two graphs create 2 right triangles on both sides of the origin, and allows us to use the Pythagorean Theorem to find half the length of the rectangle, which we can then double to find the entire length.

Notice that the length of the hypotenuse of these triangles corresponds to the radius of the circle given by the equation x² + y² = 41. Now we can solve for the unknown leg of the triangle corresponding to half the length of the rectangle because we know the other two lengths, (√2)/2 and √41 .

41 = s² + (√2/2)²
41 = s² + (√2)² / 2²
41 = s² + ½
41 - ½ = s²
(82 - 1) / 2 = s²
81/2 = s²
√(81/2) = 9 /√2 = (9√2)/2 = (9/2)√2 = s

Since s represents only half the length of the rectangle, then we double it to obtain the entire length of the rectangle:

2s = 2 [(9/2)√2] = 9√2 = L.

Since the area of a rectangle is given by LW, we can now substitute our calculated values into this equation to find the area of the elusive rectangle.

LW = (9√2)(√2) = 9(√2*2) = 9(√4) = 9(2) = 18.

So, the area of the rectangle described by this problem is 18 square units.

Answer 2

ask : what can I multiply together to get 20, and also add together their squares to get 41? well...using whole numbers... you only have 1,20 2,10 and 4,5 for your choices...

4 times 5 = 20 and 4^2 + 5^2 = 41

also -4^2 + -5^2 =41

16 + 25 = 41

so x=4 and y=5 and also they can be x=-4 and y=-5
(or vice-versa) then the points of the graph are joined together to form a rectangle. Okay so you use both choices.

And the sides of the rectangle are: (2) that are 4 and (2) that are 5.

Area of a rectangle is xy ....I think... so... ans is 20...

Oooh, nasty teacher gave you the answer in the question!