Answer:
Cost of material is $245.31
Step-by-step explanation:
Let dimension of box,
<em>The length of this base is twice the width.</em>
<em>∴ </em>x = 2y
Volume of box = 10 m³
∴ xyz = 10
⇒ 2y²z = 10
⇒ y²z = 5
![\Rightarrow z=\dfrac{5}{y^2}\ \ \ ...(i)](https://tex.z-dn.net/?f=%5CRightarrow%20z%3D%5Cdfrac%7B5%7D%7By%5E2%7D%5C%20%5C%20%5C%20...%28i%29)
<em>Material for the base costs $15 per square meter.</em>
Total cost of base = 15xy
Total cost of base = 30y² [∵ x = 2y ]
<em>Material for the sides costs $9 per square meter.</em>
Total cost of side = 9(2xz+2zy)
Total cost of side = 18(xz+yz)
Total cost of material for container = 30y² + 18(xz+yz)
[From (i)]
![C(y)=30y^2+\dfrac{270}{y}](https://tex.z-dn.net/?f=C%28y%29%3D30y%5E2%2B%5Cdfrac%7B270%7D%7By%7D)
Differentiate w.r.t y
![C'(y)=60y-\dfrac{270}{y^2}](https://tex.z-dn.net/?f=C%27%28y%29%3D60y-%5Cdfrac%7B270%7D%7By%5E2%7D)
For critical point , C'(y)=0
![60y-\dfrac{270}{y^2}=](https://tex.z-dn.net/?f=60y-%5Cdfrac%7B270%7D%7By%5E2%7D%3D)
![y=\sqrt[3]{\dfrac{9}{2}}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B9%7D%7B2%7D%7D)
![x=2y=2\sqrt[3]{\dfrac{9}{2}}](https://tex.z-dn.net/?f=x%3D2y%3D2%5Csqrt%5B3%5D%7B%5Cdfrac%7B9%7D%7B2%7D%7D)
![z=\dfrac{5}{y^2}=5\sqrt[3]{\dfrac{4}{81}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B5%7D%7By%5E2%7D%3D5%5Csqrt%5B3%5D%7B%5Cdfrac%7B4%7D%7B81%7D%7D)
The minimum cost of container material at ![y=\sqrt[3]{\dfrac{9}{2}}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B9%7D%7B2%7D%7D)
![C_{min}=30\sqrt[3]{\dfrac{81}{4}}+270\sqrt[3]{\dfrac{2}{9}}](https://tex.z-dn.net/?f=C_%7Bmin%7D%3D30%5Csqrt%5B3%5D%7B%5Cdfrac%7B81%7D%7B4%7D%7D%2B270%5Csqrt%5B3%5D%7B%5Cdfrac%7B2%7D%7B9%7D%7D)
![C_{min}=245.31](https://tex.z-dn.net/?f=C_%7Bmin%7D%3D245.31)
Hence, the cheapest cost of material for container is $245.31
Answer:
1 3/5
Step-by-step explanation:
brainliest pleaseeee
Hello!
The most logical equation would be b)
Hope it helps, have a nice day!
how many 3 element subsets of {1, 2, 3, 4, 5, 6, 7, 8, ,9, 10, 11} are there for which the sum of the elements in the subset is
AURORKA [14]
Answer:
There are 155 ways in which these elements casn occur.
Step-by-step explanation:
We want 3 element subsets whose sum are multiples of 3
1+2+3= 6
1+2+6= 9
1+2+9= 12
1+9+11=21
1+3+5=9
1+4+8=12
1+5+6=12
1+6+8=15
1+7+10=18
1+8+9=18
1+9+11=21
2+3+7=12
2+4+6=12
2+4+9=15
2+5+11=18
2+6+7=15
2+7+9=18
2+8+5=15
2+8+11=21
2+9+10=21
3+6+9= 18
3+9+11=21
3+10+11=24
6+9+10=27
6+8+11=27
6+7+11=24
7+8+9= 24
8+9+10=27
7+9+11=27 .........
We have 11 elements
We need a combination of 3
The combinations can be in the form
even+ even+ odd
odd+odd+odd
even + odd+odd
So there are 3 ways in which these elements can occur
Total number of combinations with 3 elements =11C3= 165
There are 6 odd numbers and 5 even numbers.
Number of subsets with 3 odd numbers = 6C3= 20
Number of two even numbers and 1 odd number = 5C2*6C1=10*6= 60
Number of 2 odd and 1 even number = 6C2* 5C1= 5*15= 75
So 20+60+75=155
There are 155 ways in which this combination can occur
g(x)=√(x+1)
Let's plug 10 in for x
g(10) = √(10+1)
Simplify the exponents
g(10) = √(11)
This is the exact form but it can also be expressed as a decimal.
√(11) = 3.31662479
You can simplify as needed