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**Hint:**To find the minimum daily inspection cost, we use a linear programming method of finding optimal values through constraints. Now to find the answer we need three things: Decision Variables, Objective Functions and Constraints.

The decision variables are unknown variables not given but assumed in the question, are the total number of grade I and II inspectors assigned daily.

The Objective Function is an equation formed by all the money paid and used by the two inspectors during inspection in form two variable equations.

Constraints are three or more than three equations formed by two or one variable equations and then tracing all the equations to graph and then to find the optimal value.

After finding these above three things, we need to find the feasible points from the constraints and also find the minimum optimal value using graphs.

**Complete step-by-step solution:**

The total number of grades I and II inspectors assigned by the company for inspection on a daily basis is given as: \[a.b\] respectively.

The costs processed by the company are two:

1. Payment to the Inspector.

Payment of Inspector grade I is \[Rs.5\] per hour.

Payment of Inspector grade II is \[Rs.4\] per hour.

2. Cost of inspecting errors.

Error% by first inspector \[=\dfrac{100-\text{Accuracy}%}{100}\]

Error% by first inspector \[=\dfrac{100-\text{96}%}{100}\]

\[=0.04\].

Error% by second inspector \[=\dfrac{100-\text{Accuracy}%}{100}\]

Error% by second inspector \[=\dfrac{100-\text{92}%}{100}\]

\[=0.08\].

An error made by the inspector cost company as: Error% \[\times \] Error Cost \[\times \] Pieces inspected per hour.

For Grade I inspector, the cost due to error is: \[3\times 0.04\times 20\].

For Grade II inspectors, the cost due to error is: \[4\times 0.08\times 14\].

Hence, daily cost in inspection by inspector I is the sum of Payment of Inspector I per hour and cost due to error which is \[Rs.5+Rs.\left( 3\times 0.04\times 20 \right)=Rs.7.40\].

And for Inspector II, the total cost is \[Rs.4+Rs.\left( 4\times 0.08\times 14 \right)=Rs.7.36\].

Therefore, the minimum cost in daily inspection for one hour of \[a,b\] inspectors are:

\[7.40a+7.36b\].

And for eight hours is:

\[8\left( 7.40a+7.36b \right)=59.20a+58.88b\].

\[\therefore \] The objective function is Minimize Z: \[59.20a+58.88b\].

And, the Constraints of the inspections are:

Number of grade I inspectors: \[a\le 10\] (Red)

Number of grade I inspectors: \[b\le 15\] (Blue)

And the number of inspected daily is given as:

Number of pieces \[\times \] Time of inspection \[\times \] Number of inspector \[\ge \] number of pieces inspected in total.

For both the inspectors: \[20\times 8a+14\times 8b\ge 1500\] (Green).

\[160a+112b\ge 1500\]

Now to find the feasible points we mark the equations in the graph and find the points of intersection of all three graphs.

Hence, the feasible points are:

\[\left( -1.125,15 \right)\text{ },\text{ }\left( 10,15 \right)\text{ },\text{ }\left( 10,0.893 \right)\].

Placing the values in the equation \[59.20a+58.88b\], we get (Approximate only):

\[59.20a+58.88b\]

For, \[\left( -1.125,15 \right)\], we have \[59.20a+58.88b=59.20\times -1.125+58.88\times 15=817\]

For, \[\left( 10,15 \right)\], we have \[59.20a+58.88b=59.20\times 10+58.88\times 15=1475\]

For, \[\left( 10,-0.893 \right)\], we have \[59.20a+58.88b=59.20\times 10+58.88\times -0.893=539\]

**Hence, the optimal solution or least feasible point that minimizes the daily inspection cost is \[Rs.539\].**

**Note:**Students may go wrong in many ways the first way is during finding the cost the cost of company and payment to inspector both are required . Secondly, the constraints for the number of inspectors will be less than and the cost will be greater than as we need to find how much less the company can gain profit.

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