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The number line which represents the solution set for inequality 3(8-4x)<6(x-5) is (3,inf).
Given an inequality 3(8-4x)<6(x-5).
We have to find the solution set for the given inequality.
<h3>What is inequality?</h3>
A statement of an order relationship greater than, greater than or equal to, less than, or less than or equal to between two numbers or algebraic expressions.
⇒ 24-12x<6x-30
⇒ 24+30<6x+12x
⇒ 54<18x
⇒ x>54/18
⇒ x>3
Hence, the number line which represents the solution set for inequality 3(8-4x)<6(x-5) is (3,inf).
That is starting from 3 and goes up to infinity.
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Complete Question:
A soft-drink vendor at a popular beach analyzes his sales records, and finds that if he sells x cans of soda pop in one day, his profit (in dollars) is given by P(x) = -0.001x² + 3x - 1800.
a. What is his maximum profit per day?
b. How many cans must be sold in order to obtain the maximum profit?
Answer:
a. $450
b. 1500 cans
Step-by-step explanation:
Given the following quadratic function;
P(x) = -0.001x² + 3x - 1800 ......equation 1
a. To find his maximum profit per day;
Since P(x) is a quadratic equation, P(x) would be maximum when
Note : the standard form of a quadratic equation is ax² + bx + c = 0 ......equation 2
Comparing eqn 1 and eqn 2, we have;
a = -0.001, b = 3 and c = -1800
Now, we determine the maximum profit;
Substituting the values, we have;
Cancelling out the negative signs, we have;
x at maximum = 1500
Substituting the value of "x" into equation 1;
P(1500) = -0.001 * 1500² + 3(1500) - 1800
P(1500) = -0.001 * 2250000 + 4500 - 1800
P(1500) = -2250 + 2700
P(1500) = $450
b. Therefore, the soft-drink vendor must sell 1500 cans in order to obtain the maximum profit.