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The cost (in hundreds of dollars) for Jake's Snowplowing business can be modeled by the quadratic function C(t)=−t 2 +12t, where t is the time in months. Let t=0 correspond to the start of a new year. (a) In what month does the snowplowing business have the most cost? Why do you think that is? (b) What is the most the company will have to spend to stay in business? (c) In March, what will the costs be to run the business? (d) In which month(s) will the cost be $3,200 ?

User Bluevector
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Final answer:

The month with the most cost for Jake's Snowplowing business is June. The most the company will have to spend to stay in business is $3,600. In March, the cost to run the business will be $2,700. The cost will be $3,200 in the 40th and 80th months.

Step-by-step explanation:

To find the month with the most cost for Jake's Snowplowing business, we need to determine the vertex of the quadratic function C(t) = -t^2 + 12t. The vertex can be found using the formula t = -b/2a, where a = -1 and b = 12. Plugging these values into the formula, we get t = -12/(-2) = 6.

Since t represents the time in months, the snowplowing business will have the most cost in the 6th month, which corresponds to June. This is because the quadratic function is a downward-opening parabola, and the vertex represents the lowest or highest point on the graph.

Next, to find the most the company will have to spend to stay in business, we need to calculate the maximum value of the quadratic function. We can find this by substituting the value of t = 6 into the function C(t). C(6) = -6^2 + 12(6) = -36 + 72 = 36.

Therefore, the most the company will have to spend to stay in business is $3,600 (36 hundreds of dollars).

In March, which corresponds to t = 3, we can find the cost to run the business by substituting t = 3 into the function C(t). C(3) = -3^2 + 12(3) = -9 + 36 = 27.

So in March, the cost to run the business will be $2,700 (27 hundreds of dollars).

To find the month(s) when the cost is $3,200, we need to set the quadratic function equal to 3,200 and solve for t. -t^2 + 12t = 3,200. Rearranging, we have t^2 - 12t + 3,200 = 0. By factoring or using the quadratic formula, we can find that t = 80 or t = 40.

Therefore, the cost will be $3,200 in the 40th and 80th months.

User Furbeenator
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