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
Answer:
y-8=-3/2(x+8)
Step-by-step explanation:
y-y1=m(x-x1)
y-8=-3/2(x-(-8))
y-8=-3/2(x+8)
Let's solve your system by substitution.
2
x
−
3
y
=
1
;
3
x
+
2
y
=
6
Step: Solve
2
x
−
3
y
=
1
for x:
2
x
−
3
y
=
1
2
x
−
3
y
+
3
y
=
1
+
3
y
(Add 3y to both sides)
2
x
=
3
y
+
1
2
x
2
=
3
y
+
1
2
(Divide both sides by 2)
x
=
3
2
y
+
1
2
Step: Substitute
3
2
y
+
1
2
for
x
in
3
x
+
2
y
=
6
:
3
x
+
2
y
=
6
3
(
3
2
y
+
1
2
)
+
2
y
=
6
13
2
y
+
3
2
=
6
(Simplify both sides of the equation)
13
2
y
+
3
2
+
−
3
2
=
6
+
−
3
2
(Add (-3)/2 to both sides)
13
2
y
=
9
2
13
2
y
13
2
=
9
2
13
2
(Divide both sides by 13/2)
y
=
9
13
Step: Substitute
9
13
for
y
in
x
=
3
2
y
+
1
2
:
x
=
3
2
y
+
1
2
x
=
3
2
(
9
13
)
+
1
2
x
=
20
13
(Simplify both sides of the equation)
Answer:
x
=
20
13
and
y
=
9
13
Answer:
52
Step-by-step explanation:
You already said that DE=52.
