What is the ratio of feet to inches on a measuring tape
1 answer:
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Answer:
a) x = 25, y = 50
b) x = 1.5, y = 3
Step-by-step explanation:
We have to use Lagrange Multipliers to solve this problem. The maximum of a differentiable function f with the constraint g(x,y) = b, then we have that there exists a constant such that
Or, in other words,
a) Lets compute the partial derivates of f(x,y) = 2x+2xy+y. Recall that, for example, the partial derivate of f respect to the variable x is obtained from derivating f thinking the variable y as a constant.
On the other hand,
The restriction is g(x,y) = 100, with g(x,y) = 2x+y. The partial derivates of g are
This means that the Lagrange equations are
(this is the restriction, in other words, g(x,y) = 100)
Note that 2y + 2, which is is the double of 2x+1, which is
. Therefore, we can forget
for now and focus on x and y with this relation:
2y+2 = 2 (2x+1) = 4x+2
2y = 4x
y = 2x
If y is equal to 2x, then
g(x,y) = 2x+y = 2x+2x = 4x
Since g(x,y) = 100, we have that
4x = 100
x = 100/4 = 25
And, therefore y = 25*2 = 50
Therefore, x = 25, Y = 50.
b) We will use the suggestion and find the minumum of f(x,y) = B² = x²+y², under the constraing g(x,y) = 0, with g(x,y) = 2x+4y-15. The suggestion is based on the fact that B is positive fon any x and y; and if 2 numbers a, b are positive, and a < b, then a² < b². In other words, if (x,y) is the minimum of B, then (x,y) is also the minimum of B² = f.
Lets apply Lagrange multipliers again. First, we need to compute the partial derivates of f:
And now, the partial derivates of g:
This gives us the following equations:
If we compare 2x with 2y, we will find that 2y is the double of 2x, because 2y is equal to , while on the other hand,
. As a consequence, we have
2y = 2*2x
y = 2x
Now, we replace y with 2x in the equation of g:
0 = g(x,y) = 2x+4y-15 = 2x+4*2x -1x = 10x-15
10 x = 15
x = 15/10 = 1.5
y = 1x5*2 = 3
Then, B is minimized for x 0 1.5, y = 3.
Answer:
Step-by-step explanation:
From the figure attached,
Point B has been dilated to form point B'.
B(3, 1) → B'(6, 2)
→ B'[(2 × 3), (2 × 1)]
Since rule for the dilation of a point (x, y) by a factor of k is,
B(x, y) → B'(kx, ky)
By comparing the coordinates k = 2 is the scale factor by which the point B has been dilated about the origin.
Therefore, other vertices of the quadrilateral will be,
A(-2, 3) → A'(-4, 6)
C(1, -1) → C'(2, -2)
D(-3, -2) → D'(-6, -4)
The distances from the point the plane leaves the ground are given by the
trigonometric relationships of right triangle and Pythagoras theorem.
- A) The minimum distance to the base of the tower is approximately <u>874.57 ft</u>.
- B) The minimum distance to the top of the tower is approximately <u>882.76 ft</u>.
Reasons:
A) The angle with which the airplane climbs, θ = 11°
Height of the tower which the airplane flies, T = 120 foot
The clearance between the tower and the airplane, C = 50 feet
Required:
The minimum distance between the point where the plane leaves the ground and the base of the tower,
Solution:
Height at which the plane flies over the tower, h = T + C
Therefore, h = 120 ft. + 50 ft. = 170 ft.
At the point the plane leaves the ground, we have;
Which gives;
- The minimum distance between the point where the plane leaves the ground and the base of the tower,
≈ <u>874.57 ft</u>.
B) The minimum distance between the point where the plane leaves the ground and the tower, <em>R</em>, is given by Pythagoras's theorem as follows;
R² = ² + T²
Which gives;
R = √(*874.57 ft.)² + (120 ft.²)) ≈ 882.76 ft.
- The distance from the point where the airplane leaves the ground to the tower, R ≈ <u>882.76 ft</u>.
Learn more about Pythagoras theorem here: