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The cost \(C\) in dollars of producing \(x\) yards of a certain fabric is given by the function $$C(x)=1500+3 x+0.02 x^{2}+0.0001 x^{3}$$ (a) Find \(C(10)\) and \(C(100)\). (b) What do your answers in part (a) represent? (c) Find \(C(0)\). (This number represents the fixed costs.)

Short Answer

Expert verified
(a) \(C(10) = 1532.1\) dollars, \(C(100) = 2100.0\) dollars; (b) Cost to produce 10 and 100 yards. (c) \(C(0) = 1500\) dollars (fixed costs).

Step by step solution

01

Substitute x in C(x) with 10

To find the cost of producing 10 yards, substitute \(x = 10\) into the function \(C(x) = 1500 + 3x + 0.02x^2 + 0.0001x^3\). This yields: \[C(10) = 1500 + 3(10) + 0.02(10)^2 + 0.0001(10)^3\]Calculate each term: - \(3(10) = 30\)- \(0.02(10)^2 = 0.02(100) = 2\)- \(0.0001(10)^3 = 0.0001(1000) = 0.1\)Finally, sum all terms: \(C(10) = 1500 + 30 + 2 + 0.1 = 1532.1\).
02

Substitute x in C(x) with 100

To find the cost of producing 100 yards, substitute \(x = 100\) into the function \(C(x) = 1500 + 3x + 0.02x^2 + 0.0001x^3\). This yields: \[C(100) = 1500 + 3(100) + 0.02(100)^2 + 0.0001(100)^3\]Calculate each term: - \(3(100) = 300\)- \(0.02(100)^2 = 0.02(10000) = 200\)- \(0.0001(100)^3 = 0.0001(1000000) = 100\)Finally, sum all terms: \(C(100) = 1500 + 300 + 200 + 100 = 2100.0\).
03

Interpret the Calculated Costs

In part (a), \(C(10) = 1532.1\) represents the cost in dollars of producing 10 yards of fabric, and \(C(100) = 2100.0\) represents the cost in dollars of producing 100 yards of fabric.
04

Find Fixed Costs by Substituting x with 0

To find the fixed costs, substitute \(x = 0\) into the function \(C(x) = 1500 + 3x + 0.02x^2 + 0.0001x^3\). This yields: \[C(0) = 1500 + 3(0) + 0.02(0)^2 + 0.0001(0)^3\]Calculate each term: - \(3(0) = 0\)- \(0.02(0)^2 = 0\)- \(0.0001(0)^3 = 0\)Then, \(C(0) = 1500\). This value represents the fixed costs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Functions
Quadratic functions are a type of polynomial function characterized by the presence of an \( x^2 \) term. In a standard form, they are expressed as \( ax^2 + bx + c \).Understanding the properties of quadratic functions can help solve a wide array of real-world problems, like predicting costs or profits.- When graphed, a quadratic function forms a parabola, which can either open upwards or downwards, depending on the sign of the \( a \) coefficient.- The vertex of the parabola (either the highest or lowest point) provides crucial insights into the function's minimum or maximum values.- In the cost function example, \( C(x) = 1500 + 3x + 0.02x^2 + 0.0001x^3 \), the \( 0.02x^2 \) represents the quadratic component influencing how costs change at a higher rate over production increases.It is important to note that each term in the function represents different influences in cost: linear (\( 3x \)), quadratic (\( 0.02x^2 \)), and cubic (\( 0.0001x^3 \)). These factors model the complexity of real-world cost scenarios.
Calculus
Calculus is a branch of mathematics focused on change and motion, utilizing concepts of derivatives and integrals to analyze functions. Derivatives, in particular, can be used to understand rates of change or to find the slope of a function at any given point. - In the context of a cost function, calculus helps determine how costs are changing with changes in production levels (i.e., finding sensitivities). - For instance, differentiating the cost function helps identify the cost's rate of increase at different production levels. - Understanding these changes allows businesses to make informed decisions about scaling production efficiently. In practice, using calculus to optimize cost functions can further help in determining break-even points or optimal production levels by examining where cost increases start to level off or significantly outpace profits. This type of analysis is crucial in setting price strategies or budgeting for expansion.
Fixed Costs
Fixed costs are expenses that do not change with the level of output. They are present even when production is zero.In our example, substituting \( x = 0 \) into the cost function reveals the fixed cost as the constant term, \( C(0) = 1500 \).- These are essential for calculating the overall cost of production as they remain constant regardless of the production volume.- Fixed costs often include elements such as rent, salaries, or equipment costs, which need to be covered by revenue before profits can be realized.- Knowing fixed costs is crucial for setting a business's budget and understanding its baseline financial requirements.By identifying and managing fixed costs effectively, companies can better predict their financial performance and minimize risks associated with variable production or market downturns.This understanding is vital for strategic planning and ensuring long-term sustainable operations.

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Most popular questions from this chapter

In a certain state the maximum speed permitted on freeways is \(65 \mathrm{mi} / \mathrm{h}\) and the minimum is \(40 .\) The fine \(F\) for violating these limits is \(\$ 15\) for every mile above the maximum or below the minimum. (a) Complete the expressions in the following piecewise defined function, where \(x\) is the speed at which you are driving. $$F(x)=\left\\{\begin{array}{ll}\text {\(\text{________ }\) if } 0 < x < 40 \\\\\text {\(\text{________ }\) if } 40 \leq x \leq 65 \\\\\text {\(\text{________ }\) if } x > 65\end{array}\right.$$ (b) Find \(F(30), F(50),\) and \(F(75)\). (c) What do your answers in part (b) represent?

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In Exercise 65 of Section 2.7 you were asked to solve equations in which the unknowns were functions. Now that we know about inverses and the identity function (see Exercise \(82),\) we can use algebra to solve such equations. For instance, to solve \(f \circ g=h\) for the unknown function \(f\) we perform the following steps: \(f \circ g=h\) Problem: Solve for \(f\) \(f \circ g \circ g^{-1}=h \circ g^{-1} \quad\) Compose with \(g^{-1}\) on the right \(f \circ I=h \circ g^{-1} \quad g \circ g^{-1}=I\) \(f=h \circ g^{-1} \quad\) f \(\circ I=f\) So the solution is \(f=h \circ g^{-1} .\) Use this technique to solve the equation \(f \circ g=h\) for the indicated unknown function. (a) Solve for \(f,\) where \(g(x)=2 x+1\) and \(h(x)=4 x^{2}+4 x+7\) (b) Solve for \(g,\) where \(f(x)=3 x+5\) and \(h(x)=3 x^{2}+3 x+2\)

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