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Flashcards: Optimization Models

A cow that weighs 175 pounds gains 3 pounds per day, and it costs 30 cents per day to keep that cow. The market price for cows is 75 cents per pound, but the price is falling one cent per day. When should the cow be sold to make the maximum profit?

Use the 5-step method.

The cow should be sold after 4 days.

The cow should be sold after 2 days.

The cow should be sold after 3 days.

The max profit is $151.58.

The max profit is $131.58.

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Flashcards: Optimization Models

Use the 5-step method.

Your answer:

The cow should be sold after 4 days.

Your answer:

The cow should be sold after 2 days.

Your answer:

The max profit is $151.58.

Your answer:

The max profit is $131.58.

Correct answer:

The cow should be sold after 3 days.

Explanation:

The 5-step method for this particular modeling question is as follows.

1. Ask the question

When should the cow be sold to make the maximum profit?

Now identify all known information.

Variables:

$\\t=\text{time (days)} \\w=\text{weight (lbs)} \\p=\text{proft (}\$\text{/lbs)} \\C=\text{cost to keep the cow} \\R=\text{revenue} \\P=\text{profit}$

$\\w=175+3t \\p=0.75-0.01t \\C=0.30t \\R=p\cdot w \\P=R-C t\geq0$

2. Select the modeling approach

For this particular question the function can be define y=f(x) on a subset of the real number line. Therefore, take the derivative of and set it equal to zero to calculate the maximum.

3. Formulate the model

To maximize formulate the equation in terms of .

$\\P=R-C \\P=p\cdot w-0.30t \\P=(0.75-0.01t)(175+3t)-0.30t \\P=131.25+0.5t-0.03t^2-0.30t \\P=131.25+0.2t-0.03t^2$

Now rewrite the equation in terms of y=f(x) .

f(x)=-0.03t^2+0.2t+131.25

4. Solve the model

Now to solve the model, take the derivative, set it equal to zero, and solve for .

$\\f'(x)=2(-0.03x^{2-1})+1(0.2x^{1-1}) \\f'(x)=-0.06x+0.2 \\0=-0.06x+0.2 \\-0.2=-0.06x \\\\\frac{-0.2}{-0.06}=x \\\\\frac{10}{3}=x$

5. Answer the question

The cow should be sold after 3 days.

Flashcards: Optimization Models

A company that makes computer monitors is unveiling two new products this spring, a 15 inch screen for $189 and a 19 inch screen for $269. The cost to manufacture the 15 inch screens is $85 per screen and the 19 inch screen is $99 per screen, plus $200,000 for miscellaneous costs. The number of sells for each screen type effects the average selling price. It is estimated that each set average selling price drops by two cents for each additional screen sold. The average sells price for the 15 inch screen will decrease by 0.2 cents for each 19 inch screen sold, the price of the 19 inch screen decreases by 0.3 cents for each 15 inch screen sold. What are the partial differential equations used to find how many units of each type of screens should be manufactured to optimize profit?

$\\\frac{\partial f}{\partial x_1}=104-0.2x_1-0.05x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.002x_2-0.05x_1=0$

$\\\frac{\partial f}{\partial x_1}=104-0.005x_1-0.02x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

$\\\frac{\partial f}{\partial x_1}=170-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=104-0.02x_2-0.005x_1=0$

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.005x_2-0.02x_1=0$

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Flashcards: Optimization Models

Your answer:

$\\\frac{\partial f}{\partial x_1}=104-0.2x_1-0.05x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.002x_2-0.05x_1=0$

Your answer:

$\\\frac{\partial f}{\partial x_1}=104-0.005x_1-0.02x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

Your answer:

$\\\frac{\partial f}{\partial x_1}=170-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=104-0.02x_2-0.005x_1=0$

Your answer:

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.005x_2-0.02x_1=0$

Correct answer:

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

Explanation:

First, identify all known information.

Variables:

$\\s=\text{number of 15 inch screens (per year)} \\t=\text{number of 19 inch screens (per year)} \\p=\text{selling price for 15 inch screen} \\q=\text{selling price for 19 inch screen} \\C=\text{cost to manufacture} \\R=\text{revenue} \\P=\text{profit}$

Assumptions:

$\\p=189-0.01s-0.002t \\q=269-0.003s-0.01t \\R=ps+qt \\C-200,000+85s+99t \\P=R-C \\s\geq0 \\t\geq0$

Objective:

Maximize the profit.

Theorem: If is a function on which is a subset of $\mathbb{R}^n$ and has a minimum or maximum at an interior point in then $\bigtriangledown f=0$ .

Therefore, to find the maximum point in this function, simultaneously solve the differential equations in question and check the boundary points.

$\\P=R-C$

$\\=ps+qt-(200,000+85s+99t) \\=(189-0.01s-0.002t)(s)+(269-0.003s -0.01t)(t)-200,000-85s-99t$

Now switch variables/

$\\y=f(x_1,..,x_n) \\x_1=s \\x_2=t$

$\\=(189-0.01x_1-0.002x_2)(x_1)+(269-0.003x_1 -0.01x_2)(x_2)\\-200,000-85x_1-99x_2$

Over

$s=\begin{Bmatrix} (x_1,x_2): x_1\geq0, x_2\geq 0 \end{Bmatrix}$

From here take the partial differential equation of each variable. Since the questioned asked for the partial differential equations that would be used to optimize the profit we can stop here.

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

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Flashcards: Optimization Models

Flashcards: Optimization Models

Flashcards: Optimization Models

Flashcards: Optimization Models

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All Math Modeling Resources

Flashcards: Optimization Models

Flashcards: Optimization Models

Flashcards: Optimization Models

Flashcards: Optimization Models

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