Math Problem Statement

Find the exact extreme values of the function z = f(x, y) = 43 x + 14 y + 70 subject to the following constraints: y >= 0 y <= 16 - x^2 Complete the following: f_(min) = -102 at (x,y) = (-4,0)

f_(max) =

Solution

To find the maximum of the function z=f(x,y)=43x+14y+70z = f(x, y) = 43x + 14y + 70 subject to the constraints y0y \geq 0 and y16x2y \leq 16 - x^2, we will proceed by checking the critical points and the boundary of the region defined by these constraints.

Step 1: Analyze the constraint y16x2y \leq 16 - x^2

The constraint y16x2y \leq 16 - x^2 defines a region bounded by the parabola y=16x2y = 16 - x^2, and since y0y \geq 0, we are only interested in the part of the parabola where 16x2016 - x^2 \geq 0, or x216x^2 \leq 16, meaning 4x4-4 \leq x \leq 4.

Thus, we are working within the following region:

y=0(the lower boundary),y=16x2(the upper boundary),4x4y = 0 \quad \text{(the lower boundary)}, \quad y = 16 - x^2 \quad \text{(the upper boundary)}, \quad -4 \leq x \leq 4

Step 2: Evaluate the function along the boundaries

Along y=0y = 0:

When y=0y = 0, the function simplifies to:

z=43x+14(0)+70=43x+70z = 43x + 14(0) + 70 = 43x + 70

Now, evaluate this for x=4x = -4 and x=4x = 4:

  • At x=4x = -4: z=43(4)+70=172+70=102z = 43(-4) + 70 = -172 + 70 = -102

  • At x=4x = 4: z=43(4)+70=172+70=242z = 43(4) + 70 = 172 + 70 = 242

Thus, on the lower boundary y=0y = 0, the function reaches a minimum value of 102-102 at x=4x = -4 and a maximum value of 242242 at x=4x = 4.

Along y=16x2y = 16 - x^2:

Now, substitute y=16x2y = 16 - x^2 into the function:

z=43x+14(16x2)+70=43x+22414x2+70=43x14x2+294z = 43x + 14(16 - x^2) + 70 = 43x + 224 - 14x^2 + 70 = 43x - 14x^2 + 294

We need to find the critical points by taking the derivative of this expression with respect to xx and setting it equal to zero:

ddx(43x14x2+294)=4328x\frac{d}{dx}(43x - 14x^2 + 294) = 43 - 28x

Setting 4328x=043 - 28x = 0, we get:

x=43281.54x = \frac{43}{28} \approx 1.54

Now, evaluate zz at x=1.54x = 1.54:

y=16(1.54)2162.37=13.63y = 16 - (1.54)^2 \approx 16 - 2.37 = 13.63

z=43(1.54)14(1.54)2+29466.2233.17+294=327.05z = 43(1.54) - 14(1.54)^2 + 294 \approx 66.22 - 33.17 + 294 = 327.05

Thus, z327.05z \approx 327.05 at x1.54x \approx 1.54.

Step 3: Conclusion

The maximum value occurs along the boundary y=16x2y = 16 - x^2 when x1.54x \approx 1.54, and the approximate maximum value is:

fmax327.05f_{\max} \approx 327.05

Summary:

  • fmin=102f_{\min} = -102 at (x,y)=(4,0)(x, y) = (-4, 0)
  • fmax327.05f_{\max} \approx 327.05 at (x,y)(1.54,13.63)(x, y) \approx (1.54, 13.63)

Would you like more details or have any questions?


Related Questions:

  1. How do we verify whether a critical point is a maximum or a minimum in constrained optimization?
  2. Can we use Lagrange multipliers for this problem? How would that work?
  3. What would the behavior of the function be if we changed the constants in the function z=43x+14y+70z = 43x + 14y + 70?
  4. How do we interpret the constraint y16x2y \leq 16 - x^2 geometrically?
  5. Could we solve this problem using a graphical method?

Tip:

When solving optimization problems with constraints, always check the boundary conditions as the extrema often occur on these boundaries.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Constrained Optimization
Parabolic Boundary

Formulas

z = 43x + 14y + 70
y <= 16 - x^2
First derivative test: dz/dx = 43 - 28x

Theorems

Optimization on Boundary
Critical Points of Multivariable Functions

Suitable Grade Level

College-Level Calculus