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rjkz [21]
3 years ago
5

Minimizing Packaging Costs If an open box has a square base and a volume of 107 in.3 and is constructed from a tin sheet, find t

he dimensions of the box, assuming a minimum amount of material is used in its construction. (Round your answers to two decimal places.)
Mathematics
1 answer:
uysha [10]3 years ago
5 0

Answer:

The dimensions of the box that minimizes the amount of material of construction is

Square base = (5.98 × 5.98) in²

Height of the box = 2.99 in.

Step-by-step explanation:

Let the length, breadth and height of the box be x, z and y respectively.

Volume of the box = xyz = 107 in³

The box has a square base and an open top.

x = z

V = x²y = 107 in³

The task is to minimize the amount of material used in its construction, that is, minimize the surface area of the box.

Surface area of the box (open at the top) = xz + 2xy + 2yz

But x = z

S = x² + 2xy + 2xy = x² + 4xy

We're to minimize this function subject to the constraint that

x²y = 107

The constraint can be rewritten as

x²y - 107 = constraint

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x and y

L(x,y,z) = x² + 4xy - λ(x²y - 107)

We then take the partial derivatives of the Lagrange function with respect to x, y and λ. Because these are turning points, at the turning points each of the partial derivatives is equal to 0.

(∂L/∂x) = 2x + 4y - 2λxy = 0

λ = (2x + 4y)/2xy = (1/y) + (2/x)

(∂L/∂y) = 4x - λx² = 0

λ = (4x)/x² = (4/x)

(∂L/∂λ) = x²y - 107 = 0

We can then equate the values of λ from the first 2 partial derivatives and solve for the values of x and y

(1/y) + (2/x) = (4/x)

(1/y) = (2/x)

x = 2y

Hence, at the point where the box has minimal area,

x = 2y

Putting these into the constraint equation or the solution of the third partial derivative,

x²y - 107 = 0

(2y)²y = 107

4y³ = 107

y³ = (107/4) = 26.75

y = ∛(26.75) = 2.99 in.

x = 2y = 2 × 2.99 = 5.98 in.

Hence, the dimensions of the box that minimizes the amount of material of construction is

Square base = (5.98 × 5.98) in²

Height of the box = 2.99 in.

Hope this Helps!!!

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Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

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where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

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N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

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⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




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The number of ways to form a committee.

Solution:

We have,

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Number of ways to select r items from total n items is

^nC_r=\dfrac{n!}{r!(n-r)!}

Number of ways to select 6 members from total 100 members is

^{100}C_6=\dfrac{100!}{6!(100-6)!}

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