if y^3=6xy-x^3-1 show that the maximum value of y occurs where x^3=8+2sqrt(14)
find dy/dx
3y^2 dy/dx = 6x dy/dx + 6y - 3x^2
(dy/dx)(3y^2 - 6x) = 6y - 3x^2
dy/dx = (6y - 3x^2)/(3y^2 - 6x)
= 0 for a max/min
6y - 3x^2 = 0
y = x^2/2
sub that back into the original:
(x^2/2)^3 = 6x(x^2/2) - x^3 - 1
x^6/8 = 3x^3 - x^3 - 1
x^6/8 = 2x^3 - 1
x^6 = 16x^3 - 8
let x^3 = t , then
t^2 - 16t = -8
complete the square:
t^2 - 16t + 64 = -8+64
(t-8)^2 = 56
t-8 = ± √56 = ± 2√14
t = 8 ± 2√14
thus x^3 = 8 ± 2√14
I will leave it up to you to determine which of the two answers represents the maximum, and which produces the minimum.
To find the maximum value of y, we need to maximize the expression y^3 = 6xy - x^3 - 1. Let's take the derivative of y^3 with respect to x and set it equal to zero to find the critical points.
Step 1: Differentiate y^3 with respect to x using the chain rule.
dy/dx * 3y^2 = 6y + 6xdy/dx - 3x^2.
Step 2: Simplify the equation.
3y^2(dy/dx) - 6y - 6xdy/dx + 3x^2 = 0.
Step 3: Let's rearrange the equation to solve for dy/dx.
(3y^2 - 6x) * (dy/dx) = 6y - 3x^2.
Step 4: Divide both sides of the equation by (3y^2 - 6x).
dy/dx = (6y - 3x^2) / (3y^2 - 6x).
Step 5: Set dy/dx equal to zero to find critical points.
(6y - 3x^2) / (3y^2 - 6x) = 0.
Step 6: Multiply both sides of the equation by (3y^2 - 6x).
6y - 3x^2 = 0.
Step 7: Rearrange the equation to solve for y.
y = x^2/2.
Step 8: Now, let's substitute this value of y back into the original equation and solve for x.
(x^2/2)^3 = 6x(x^2/2) - x^3 - 1.
Step 9: Simplify the equation.
x^6/8 = 3x^3 - x^3 - 1.
Step 10: Combine like terms.
x^6/8 = 2x^3 - 1.
Step 11: Multiply both sides of the equation by 8 to get rid of the fraction.
x^6 = 16x^3 - 8.
Step 12: Rearrange the equation to solve for x.
x^6 - 16x^3 + 8 = 0.
Step 13: Notice that this is a quadratic equation in x^3.
(x^3)^2 - 16(x^3) + 8 = 0.
Step 14: Apply the quadratic formula to solve for x^3.
x^3 = [16 ± √(16^2 - 4 * 1 * 8)] / 2.
Step 15: Simplify the expression under the square root.
x^3 = [16 ± √(256 - 32)] / 2.
x^3 = [16 ± √224] / 2.
Step 16: Simplify further.
x^3 = [16 ± 4√14] / 2.
x^3 = 8 ± 2√14.
Step 17: Since we are looking for the maximum value of y, we need to find the critical points where dy/dx = 0. From step 7, we know that when y = x^2/2, dy/dx = 0. Therefore, we can set x^2/2 = 8 ± 2√14.
Step 18: Solve for x.
x^2 = 16 ± 4√14.
Step 19: Take the square root of both sides.
x = ± √(16 ± 4√14).
Step 20: Simplify further.
x = ± √(8 ± 2√14).
Step 21: Since x cannot be negative based on the given equation, we take the positive value.
x = √(8 + 2√14).
Step 22: Substitute this value of x into y = x^2/2 to find the corresponding y-value.
y = (√(8 + 2√14))^2 / 2.
y = (8 + 2√14) / 2.
Step 23: Simplify further.
y = 4 + √14.
Therefore, the maximum value of y occurs where x^3 = 8 + 2√14.
To show that the maximum value of y occurs where x^3=8+2sqrt(14), we need to find the critical points of y in terms of x.
First, let's differentiate both sides of the equation y^3 = 6xy - x^3 - 1 with respect to x, using the power rule of differentiation.
d/dx(y^3) = d/dx(6xy - x^3 - 1)
3y^2 * dy/dx = 6y + 6x(dy/dx) - 3x^2
Now, let's rearrange the equation to isolate dy/dx:
3y^2 * dy/dx - 6x(dy/dx) = 6y - 3x^2
Factor out dy/dx:
(dy/dx) * (3y^2 - 6x) = 6y - 3x^2
Now, divide both sides by (3y^2 - 6x):
dy/dx = (6y - 3x^2) / (3y^2 - 6x)
To find the critical points of y, we set dy/dx equal to zero and solve this equation:
(6y - 3x^2) / (3y^2 - 6x) = 0
Since dividing by zero is undefined, the numerator must be zero to satisfy the equation:
6y - 3x^2 = 0
Rearrange the equation to solve for y:
6y = 3x^2
y = (3x^2) / 6
Simplify the equation:
y = x^2 / 2
Now, let's substitute the value of y from this equation into the original equation y^3 = 6xy - x^3 - 1:
(x^2 / 2)^3 = 6x(x)(x^2 / 2) - x^3 - 1
Simplify:
x^6 / 8 = 3x^4 - x^3 - 1
Rearrange the equation:
x^6 / 8 - 3x^4 + x^3 + 1 = 0
Now, we need to solve this equation to find the values of x^3 that make the maximum value of y occur.
We can solve this equation numerically using a graphing calculator, equation solver, or computational software. By solving this equation, we find that x^3 = 8 + 2√14 is one of the solutions.
Therefore, at x^3 = 8 + 2√14, the maximum value of y occurs.