The daily revenue R, in dollars, of a manufacturing company is modeled by the quadratic fu… — ACT

The daily revenue $R$ , in dollars, of a manufacturing company is modeled by the quadratic function $R(x) = -0.2x^2 + kx - 1,200$ , where $x$ is the number of units produced and sold, and $k$ is a positive constant. If the maximum daily revenue the company can achieve is $800$ dollars, what is the value of $k$ ?

Answer: 40

Answer

The value of the constant

k

is 40.

Setting the daily revenue function equal to 800 and rewriting it in standard form yields

-0.2x^2 + kx - 2000 = 0

. For a quadratic equation to have exactly one real solution, which represents the maximum vertex of the parabola, the discriminant must be equal to 0. Setting the discriminant

b^2 - 4ac = 0

gives

k^2 - 4(-0.2)(-2000) = 0

, which simplifies to

k^2 - 1600 = 0

. Solving for the positive constant

k

gives

k = 40

Step-by-Step Solution

Set the revenue function equal to the maximum daily revenue of 800 dollars.

-0.2x^2 + kx - 1,200 = 800

The maximum revenue is the highest point (vertex) on the parabola, where the line

y = 800

is tangent to the curve.

Convert the equation into standard quadratic form

ax^2 + bx + c = 0

-0.2x^2 + kx - 2,000 = 0

Standard form is required to identify the coefficients

a

b

, and

c

for the discriminant formula.

Set the discriminant equal to zero.

k^2 - 4(-0.2)(-2,000) = 0

Since the maximum daily revenue is achieved at exactly one point, the quadratic equation must have exactly one real solution, meaning its discriminant (

b^2 - 4ac

) must be zero.

Solve for the positive constant

k

k^2 - 1,600 = 0 \implies k = 40

Solving the equation yields

k = \pm 40

. Since the problem specifies that

k

is a positive constant, we select

k = 40

Key Concept

Using the discriminant of a quadratic equation to find the value of a parameter when there is exactly one real solution.

Estimated Time:2m 0s

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