Math Problem Statement

A chef is estimating the cost to prepare a fixed-price menu for x guests. The anticipated cost of the appetizers is a(x) = 3.25x, the anticipated cost for the main course is m(x) = 5.25x, and the anticipated cost for the dessert course is d(x) = 1.25x. If the functions are combined so that c(x) = a(x) + m(x) + d(x), then the function c(x) includes the point (40,

Solution

To find the function c(x)c(x) that includes the point (40,y)(40, y), we first need to combine the given functions a(x)a(x), m(x)m(x), and d(x)d(x).

Given:

  • a(x)=3.25xa(x) = 3.25x (cost of appetizers)
  • m(x)=5.25xm(x) = 5.25x (cost of the main course)
  • d(x)=1.25xd(x) = 1.25x (cost of dessert)

The total cost function c(x)c(x) is the sum of these three functions:

c(x)=a(x)+m(x)+d(x)c(x) = a(x) + m(x) + d(x)

Substituting the expressions for a(x)a(x), m(x)m(x), and d(x)d(x):

c(x)=3.25x+5.25x+1.25xc(x) = 3.25x + 5.25x + 1.25x

Now, combine like terms:

c(x)=(3.25+5.25+1.25)x=9.75xc(x) = (3.25 + 5.25 + 1.25)x = 9.75x

So, the function c(x)c(x) is:

c(x)=9.75xc(x) = 9.75x

We are given that the function c(x)c(x) includes the point (40,y)(40, y). To find yy, substitute x=40x = 40 into the function:

y=c(40)=9.75×40y = c(40) = 9.75 \times 40

y=390y = 390

Thus, the point is (40,390)(40, 390).

Would you like more details or have any questions? Here are five related questions:

  1. What would the cost be for 50 guests?
  2. How would the cost function change if the price of the main course increased by $2 per guest?
  3. What is the total cost for 100 guests?
  4. How would you adjust the function if the dessert cost doubled?
  5. What is the individual contribution of each course to the total cost for 40 guests?

Tip: Always check your function by substituting values to ensure the calculations are consistent with given points.

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Math Problem Analysis

Mathematical Concepts

Functions
Linear Equations
Cost Analysis

Formulas

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Theorems

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Suitable Grade Level

Grades 9-12