find the values of a and b if the function y=2x^3+ax^2+bx+36 has a local maximum when x=-4 and a local minimum when x=5

Sure, I'll try to find the values of a and b for you. Before I do that, I must say that math and humor are like clowns and tightrope walking – a bit of a balancing act. Let's proceed with caution!

To find the values of a and b, we need to consider the behavior of the function at the given points. At a local maximum or minimum, the derivative of the function should be equal to zero.

So let's take the derivative of the function y = 2x^3 + ax^2 + bx + 36:

dy/dx = 6x^2 + 2ax + b

Now, since we want a local maximum at x = -4, we can set the derivative to zero and solve for a and b:

0 = 6(-4)^2 + 2a(-4) + b
0 = 96 - 8a + b (Equation 1)

Similarly, for a local minimum at x = 5:

0 = 6(5)^2 + 2a(5) + b
0 = 150 + 10a + b (Equation 2)

Now, to solve these two equations simultaneously, we can subtract Equations 1 and 2:

(96 - 8a + b) - (150 + 10a + b) = 0
-54 - 18a = 0
-18a = 54
a = -3

Substituting this value of a into Equation 1:

96 - 8(-3) + b = 0
96 + 24 + b = 0
b = -120

So, the values of a and b that satisfy the given conditions are a = -3 and b = -120. Voila!

Remember, this clown is an expert in math humor but sometimes the actual calculations can be a bit tricky. Hope this helps!

To find the values of a and b, we need to determine the derivatives of the given function, set them equal to zero, and use the given conditions to solve for a and b.

Step 1: Find the derivative of the function y with respect to x.
y = 2x^3 + ax^2 + bx + 36
The first derivative is:
dy/dx = 6x² + 2ax + b

Step 2: Set the derivative equal to zero to find the critical points.
6x² + 2ax + b = 0

Step 3: Substitute the given x-values (-4 and 5) to create two equations.
At x = -4:
6(-4)² + 2a(-4) + b = 0
16a - 8a + b = 0
8a + b = 0 -----(Equation 1)

At x = 5:
6(5)² + 2a(5) + b = 0
50a + 10a + b = 0
60a + b = 0 -----(Equation 2)

Step 4: Solve the system of equations (Equation 1 and Equation 2).

Multiply Equation 1 by 6:
48a + 6b = 0 ----(Equation 3)

Subtract Equation 3 from Equation 2 to eliminate b:
(60a + b) - (48a + 6b) = 0
12a - 5b = 0 ----(Equation 4)

Step 5: Solve Equations 3 and 4 simultaneously.

Multiply Equation 4 by 6:
72a - 30b = 0 ----(Equation 5)

Multiply Equation 3 by 2:
96a + 12b = 0 ----(Equation 6)

Add Equation 5 and Equation 6 to eliminate b:
(72a - 30b) + (96a + 12b) = 0
168a - 18b = 0

Divide both sides by 6:
28a - 3b = 0 ----(Equation 7)

We can now set Equation 7 equal to Equation 4 and solve for a:

28a - 3b = 12a - 5b
16a = 2b
8a = b

Substituting 8a for b in Equation 7:
28a - 3(8a) = 0
28a - 24a = 0
4a = 0
a = 0

Step 6: Substitute the value of a back into Equation 1 to solve for b:

8a + b = 0
8(0) + b = 0
b = 0

Therefore, the values of a and b are both 0.

To find the values of a and b, we need to use the information given about the local maximum and minimum of the function.

First, let's find the derivative of the given function y = 2x^3 + ax^2 + bx + 36. The derivative will give us the equation for the slope of the function at different points.

Differentiating the function with respect to x:

dy/dx = 6x^2 + 2ax + b

To find the critical points (where the local maximum and minimum occur), we set the derivative equal to zero and solve for x:

6x^2 + 2ax + b = 0

Given that the local maximum occurs when x = -4, substitute this value into the equation:

6(-4)^2 + 2a(-4) + b = 0
96 - 8a + b = 0 -- (equation 1)

Similarly, given that the local minimum occurs when x = 5, substitute this value into the equation:

6(5)^2 + 2a(5) + b = 0
150 + 10a + b = 0 -- (equation 2)

We now have a system of two equations (equation 1 and equation 2) with two unknowns (a and b).

Solving the system of equations can help us find the values of a and b.

Using any preferred method of solving systems of linear equations, we can find that a = -6 and b = 6.

Therefore, the values of a and b for the function y = 2x^3 + ax^2 + bx + 36 are a = -6 and b = 6.

at both max and min, the derivative is zero, so

dy/dx = 6x^2 + 2ax + b
at x=-4,
96 - 8a + b = 0
at x=5
150 + 10a + b = 0
subtract them
54 + 16a = 0
a = - 27/8

sub that back into one of the above equations to find b