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:

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

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

2. Select the modeling approach

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

3. Formulate the model

To maximize \(\displaystyle P\) formulate the equation in terms of \(\displaystyle t\).

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

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

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

4. Solve the model

Now to solve the model, take the derivative, set it equal to zero, and solve for \(\displaystyle x\).

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

5. Answer the question

The cow should be sold after 3 days.

First, identify all known information.

Variables:

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

Assumptions:

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

Objective:

Maximize the profit.

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

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

\(\displaystyle \\P=R-C\)

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

Now switch variables/

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

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

Over 

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

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.

\(\displaystyle \\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0\)