Questions: P=-76 x^3+4810 x^2-320,000, 0 leq x leq 60 Using this model, find the smaller of two advertising amounts that will yield a profit of 2,490,000. (Round your answer to the nearest thousands of dollars.) x

P=-76 x^3+4810 x^2-320,000, 0 leq x leq 60

Using this model, find the smaller of two advertising amounts that will yield a profit of 2,490,000. (Round your answer to the nearest thousands of dollars.)
x

Transcript text: \[ P=-76 x^{3}+4810 x^{2}-320,000, \quad 0 \leq x \leq 60 \] Using this model, find the smaller of two advertising amounts that will yield a profit of $\$ 2,490,000$. (Round your answer to the nearest thousands of dollars.) $\$$ $\square$ $x$

Solution

Solution Steps

Step 1: Define the Profit Equation

The profit function is given by

\[ P = -76x^3 + 4810x^2 - 320,000 \]

We need to find the values of \( x \) such that

\[ P = 2,490,000 \]

This leads us to the equation:

\[ -76x^3 + 4810x^2 - 320,000 - 2,490,000 = 0 \]

which simplifies to

\[ -76x^3 + 4810x^2 - 2,810,000 = 0 \]

Step 2: Solve for Roots

Using numerical methods, we find the roots of the cubic equation. The calculated roots are approximately

\[ 39.0796, \quad 39.0796, \quad 45.1601 \]

Step 3: Filter Valid Roots

We need to filter these roots to find those that lie within the range \( 0 \leq x \leq 60 \). The valid roots are

\[ 0.0, \quad 0.0, \quad 0.0 \]

Step 4: Identify the Smaller Root

Among the valid roots, the smallest value is

\[ 0.0 \]

Final Answer

The smaller of the two advertising amounts that will yield a profit of \$2,490,000 is

\[ \boxed{x = 0.0} \]

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