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Tatiana [17]
3 years ago
14

A school has 147 third graders. The third grade teachers have planned a special treat for the last day of school and brought ice

cream for their students. There are three flavors: mint chip, chocolate, and strawberry. Suppose that 60 students like (at least) mint chip, 103 like chocolate, 50 like strawberry, 30 like mint chip and strawberry, 40 like mint chip and chocolate, 25 like chocolate and strawberry, and 18 like all three flavors. How many students don't like any of the flavors available
Mathematics
1 answer:
Hunter-Best [27]3 years ago
3 0
Add all of the students who like the ice creams and subtract from 147 :)
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Keegan is priting and selling his original design on t-shirts. He has concluded that for x shirts, in thousands sold his total p
Damm [24]

Complete question is;

Keegan is printing and selling his original design on t-shirts. He has concluded that for x shirts, in thousands sold his total profits will be p(x) = -x³ + 4x² + x dollars, in thousands will be earned. How many t-shirts (rounded to the nearest whole number) should he print in order to make maximum profits? What will his profits rounded to the nearest whole dollar be if he prints that number of shirts?

Answer:

Number of t-shirts to make maximum profit = 2790 shirts

Maximum profit = $12,209

Step-by-step explanation:

From the question, we are given that the profit function is;

p(x) = -x³ + 4x² + x

For the maximum value of the profit function,

(dp/dx) = 0 and (d²p/dx²) < 0

Since, p(x) = -x³ + 4x² + x

Then,

(dp/dx) = -3x² + 8x + 1

at maximum point (dp/dx) = 0, thus;

-3x² + 8x + 1 = 0

Solving this using quadratic formula, the roots are;

x = -0.12 or 2.79

Also, (d²p/dx²) = -6x + 8

Now, let's put the roots of x into -6x + 8 and check for maximum value conditon;

at x = -0.12

(d²p/dx²) = -6(0.12) + 8 = 7.28 > 0

At x = 2.79

(d²p/dx²) = -6(2.79) + 8 = -8.74 < 0

Maximum has to be d²p/dx² < 0

So, the one that meets the condition is -8.74 < 0 at x = 2.79

Thus, the maximum of the profit function exists when the number of shirts, x = 2.79 (in thousands) = 2790

Now, the maximum profits that corresponds to this number of t-shirts of 2.79(in thousands) is obtained by putting 2.79 for x in the profit function;

So,

p(2.79) = -(2.79)³ + 4(2.79²) + 2.79

p(x) = -21.7176 + 31.1364 + 2.79

p(x) = 12.2088 (in thousand dollars) ≈ $12,209

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