The tangent line is perpendicular to the radius from the center of the circle to the point of tangency.
ΔDCE is similar to ΔDBA
DC/DB=DE/DA
15/50=DE/40
DE=12
Answer:
1. 155 yd², 2. 379.54 ft², 5. x = 65.82π
Step-by-step explanation:
1. A = (1/2) · (3 + 13) · 8
A = (1/2)(16)(8)
A = 64
A = bh
A = 13*7
A = 91
64 + 91 = 155
155 yd²
2. A = bh
A = 25x25
A = 625
A = πr²
A = 3.14*12.5²
A = 3.14*156.25
A = 490.625
490.925/2 = 245.4625
625 - 245.4625 = 379.5375
379.54 ft²
3. and 4. Sorry, I don't know how to solve these
5. A = πr²
A = 3.14*7²
A = 3.14*49
A = 153.86
x/154 = 153.86π/360
360x = 154*153.86π
360x = 23694.44π
x = 65.8178889π
x = 65.82π
Hope this helps :)
The first is the first one tho
Answer:
the dimensions of the box that minimizes the cost are 5 in x 40 in x 40 in
Step-by-step explanation:
since the box has a volume V
V= x*y*z = b=8000 in³
since y=z (square face)
V= x*y² = b=8000 in³
and the cost function is
cost = cost of the square faces * area of square faces + cost of top and bottom * top and bottom areas + cost of the rectangular sides * area of the rectangular sides
C = a* 2*y² + a* 2*x*y + 15*a* 2*x*y = 2*a* y² + 32*a*x*y
to find the optimum we can use Lagrange multipliers , then we have 3 simultaneous equations:
x*y*z = b
Cx - λ*Vx = 0 → 32*a*y - λ*y² = 0 → y*( 32*a-λ*y) = 0 → y=32*a/λ
Cy - λ*Vy = 0 → (4*a*y + 32*a*x) - λ*2*x*y = 0
4*a*32/λ + 32*a*x - λ*2*x*32*a/λ = 0
128*a² /λ + 32*a*x - 64*a*x = 0
32*a*x = 128*a² /λ
x = 4*a/λ
x*y² = b
4*a/λ * (32*a/λ)² = b
(a/λ)³ *4096 = 8000 m³
(a/λ) = ∛ ( 8000 m³/4096 ) = 5/4 in
then
x = 4*a/λ = 4*5/4 in = 5 in
y=32*a/λ = 32*5/4 in = 40 in
then the box has dimensions 5 in x 40 in x 40 in
