# Tina's Material Company hauls gravel to a construction site, using a small truck and a large truck. The carrying capacity and operating cost per load are given in the accompanying table. Tina...

Tina's Material Company hauls gravel to a construction site, using a small truck and a large truck. The carrying capacity and operating cost per load are given in the accompanying table. Tina must deliver a minimum of 160 cubic yards per day to satisfy her contract with the builder. The union contract with her drivers requires that the total number of loads per day is a minimum of 9. How many loads should be made in each truck per day to minimize the total cost?

Capacity in cubic yards for a small truck are 20 and for a large truck are 80.

Cost per load for a small truck are $82 and for a large truck are $47.

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Assign the variables:

x = small truck

y = large truck

T = total cubic yards

C = cost

A small truck can hold 20 cubic yards.

A large truck can hold 80 cubic yards.

Therefore, the formula for total cubic yards is...

T = 20x + 80y

A small truck load is $82.

A large truck load is $47.

Therefore, the formula for cost is...

C = 82x + 47y

There must be a minimum of 9 loads.

Any combination of 9 small or large trucks will meet the required minimum of 160 cubic yards.

For example, 9 small trucks and 0 large trucks:

T = 20x + 80y

T = 20 * 9 + 80 * 0

T = 180

At the other extreme, 0 small trucks and 9 large trucks:

T = 20x + 80y

T = 20 * 0 + 80 * 9

T = 720

Therefore, you only have to concern yourself with minimizing the cost. Calculate the cost of each possible combination, using the formula C = 82x + 47y

(0, 9) 82 * 0 + 47 * 9 = $423

(1, 8) 82 * 1 + 47 * 8 = $458

(2, 7) 82 * 2 + 47 * 7 = $493

You can see the pattern. The more small trucks are used, the higher the cost. Therefore, you know the most cost-effective option would be to use 9 large trucks. This would provide 80 * 9 = 720 cubic yards and would cost 47 * 9 = $423.

The mathematcal model of problem is

Let small truck carries x cubic loads and large truck carries y cubic loads .

Thus

Cost `C=82x+47y`

sub to.

capacity

`20x+80y>=160`

and

union

`x+y>=9` ,`x,y>=0`

Thus

Min `C=82x+47y`

s.t.

`x+4y>=8`

`x+y>=9`

`x,y>=0`

Now draw graphs of constraints

The solution of inequality is in first quadrant right of green line.

Thus pssible solutionare P(9,0) and Q(0,9)

C at P(9,0)=82 x 9+47 x 0=738

C at (0,9)=82 x 0+47 x 9= 423

Thus minimum of C= 423 and it will at Q(0,9)

i.e x=0 ,y=9 and cost =423.