We are given the following:
To find the value of \( x \) (the number of items) that will give a profit, we need to calculate the profit function. The profit function is the difference between revenue and cost:
\[ \text{Profit} = R(x) - C(x) \]
Substituting the given functions:
\[ \text{Profit} = (60x + 2000) - (20x + 4000) \]
Simplifying:
\[ \text{Profit} = 60x + 2000 - 20x - 4000 \]
\[ \text{Profit} = 40x - 2000 \]
To earn a profit, the profit must be greater than 0:
\[ 40x - 2000 > 0 \]
Solving for \( x \):
\[ 40x > 2000 \]
\[ x > \frac{2000}{40} \]
\[ x > 50 \]
Therefore, the value of \( x \) to earn a profit is greater than 50.
The correct answer is (A): \( x > 50 \)
We are given the cost function \( C(x) \) and the revenue function \( R(x) \) for a product, where x is the number of items produced and sold:
The Profit function \( P(x) \) is calculated as the difference between the revenue and the cost:
\[ P(x) = R(x) - C(x) \]
Substitute the given functions into the profit formula:
\[ P(x) = (60x + 2000) - (20x + 4000) \]
Simplify the expression:
\[ P(x) = 60x + 2000 - 20x - 4000 \]
\[ P(x) = (60x - 20x) + (2000 - 4000) \]
\[ P(x) = 40x - 2000 \]
To earn a profit, the profit \( P(x) \) must be greater than zero:
\[ P(x) > 0 \]
Substitute the expression for \( P(x) \):
\[ 40x - 2000 > 0 \]
Now, solve this inequality for x:
Add 2000 to both sides:
\[ 40x > 2000 \]
Divide both sides by 40:
\[ x > \frac{2000}{40} \]
\[ x > \frac{200}{4} \]
\[ x > 50 \]
Therefore, a profit is earned when the number of items produced and sold (x) is greater than 50.
Comparing this result with the given options:
The correct option is >50.
Let A be the set of 30 students of class XII in a school. Let f : A -> N, N is a set of natural numbers such that function f(x) = Roll Number of student x.
Give reasons to support your answer to (i).
Find the domain of the function \( f(x) = \cos^{-1}(x^2 - 4) \).
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: