I have two problems (listed below):

1.The profit P (in hundreds of dollars) that a company makes depends on the amount x (in hundreds of dollars) the company spends on advertising according to the model P = 230 + 20x -0.5x^2.

What expenditure for advertising will yield a maximum profit?

2. The height y (in feet) of a ball thrown by a child is y= -1x^2/12 + 2x + 4, where x is the horizontal distance (in feet) from the point at which the ball is thrown.

How far from the child does the ball strike the ground?

1. Ah, the classic advertising dilemma! To find the expenditure that will yield a maximum profit, we need to maximize the quadratic equation P = 230 + 20x - 0.5x^2. Now, no need to quack your brains over this - let's throw in some math fun!

First, let's rewrite the equation as P = -0.5x^2 + 20x + 230. Remember, this is for finding the maximum, not the minimum – we're not looking to minimize profits; we're here to maximize the moolah!

To find the maximum point, we take the derivative of the equation. So, dP/dx = -1x + 20. Great, now let's set this der-ivy equal to zero: -x + 20 = 0. Solving for x, we get x = 20.

So, an expenditure of $20 (in hundreds of dollars) will maximize the profit! Just keep in mind, this is a quadratic model, not a magical profit genie – accuracy may vary!

2. Ah, the ball in flight! But where will it land? Let's dig in and find out!

We have the equation y = -1x^2/12 + 2x + 4. To figure out where the ball strikes the ground, we need to find the x-intercept. In simpler terms, we're looking for when y becomes zero – or when the ball hits the ground with a resounding thud!

Setting y to zero, we have 0 = -1x^2/12 + 2x + 4. Now, let's put on our math hat and solve this bad boy!

Multiplying everything by 12 to eliminate the fraction gives us 0 = -x^2 + 24x + 48. A bit of algebra acrobatics will reveal that x^2 - 24x - 48 = 0. Now we can use the quadratic formula for a grand finale!

The quadratic formula tells us that x = (-b ± √(b^2 - 4ac)) / 2a. Plugging in the values a = 1, b = -24, and c = -48, we find x equals approximately 25.17 or -1.17.

Since we're dealing with a distance, a negative value doesn't make sense. Therefore, the ball strikes the ground approximately 25.17 feet away from the child.

And there you have it, my friend! Playful math and a bit of clowny wisdom to solve your problems!

To find the expenditure for advertising that will yield a maximum profit, we need to use calculus and find the critical points of the profit equation, P = 230 + 20x - 0.5x^2.

To find the critical points, we need to take the derivative of P with respect to x, set it equal to zero, and solve for x.

First, let's find the derivative of P with respect to x:
dP/dx = 20 - x

Next, set dP/dx equal to zero and solve for x:
20 - x = 0
x = 20

So, x = 20 is a critical point of the profit equation.

To determine if it is a maximum or minimum, we need to take the second derivative and evaluate it at the critical point.

The second derivative, d^2P/dx^2, is the derivative of the derivative:
d^2P/dx^2 = -1

Since the second derivative is negative, it means the critical point x = 20 corresponds to a maximum profit.

Therefore, an expenditure of x = 20 (in hundreds of dollars) for advertising will yield a maximum profit.

Moving on to the second problem, to find how far from the child the ball strikes the ground, we need to determine the value of x when y = 0 in the equation y = -1x^2/12 + 2x + 4.

To do this, we set y equal to 0 and solve for x:
0 = -x^2/12 + 2x + 4

To simplify the equation, let's multiply it by 12 to clear the fractions:
0 = -x^2 + 24x + 48

Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula to find the solutions for x:

x = (-b ± √(b^2 - 4ac)) / 2a

For the quadratic equation -x^2 + 24x + 48, the coefficients are a = -1, b = 24, and c = 48.

Substituting these values into the quadratic formula:

x = (-24 ± √(24^2 - 4*(-1)*(48))) / (2*(-1))
x = (-24 ± √(576 + 192)) / (-2)
x = (-24 ± √(768)) / (-2)
x = (-24 ± 2√(192)) / (-2)
x = 12 ± √(192)

So, the ball strikes the ground when it is approximately 12 - √(192) feet and 12 + √(192) feet away from the child.