Math Problem Statement

En funktion f er bestemt ved 2 fx x () 4 =- + . Grafen for f og koordinatsystemets akser afgrænser i første kvadrant en punktmængde M, der har et areal. a) Bestem arealet af M. Når 0 2 < <a , skærer tangenten til grafen for f i punktet ( , ( )) Pa f a koordinatsystemets akser i punkterne Q og R (se figuren). Det oplyses, at arealet af trekant OQR er en funktion af a, som er givet ved 2 2 ( 4) ( ) , 0 2. 4 a Ta a a + = << b) Bestem den værdi af a, der gør arealet af trekant OQR mindst muligt. c) Bestem koordinatsættet til hvert af punkterne Q og R udtrykt ved a, og gør rede for, at arealet af trekant OQR som funktion af a er givet ved ( ).

Solution

Let's break down the problem into the steps needed to solve it:

Part a: Determine the Area of M

  1. The function is given by f(x)=2x+4f(x) = -2x + 4. To find the area enclosed by the graph of ff and the coordinate axes in the first quadrant:
    • Identify the intercepts:
      • xx-intercept: Set f(x)=0f(x) = 0: 2x+4=0x=2-2x + 4 = 0 \Rightarrow x = 2
      • yy-intercept: Set x=0x = 0: f(0)=4f(0) = 4
    • The intercepts are at points (2,0)(2, 0) and (0,4)(0, 4).
    • The area MM in the first quadrant forms a right triangle with these intercepts: Area=12×base×height=12×2×4=4\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4

Part b: Find the Value of aa Minimizing the Area of Triangle OQR

  1. Given the function T(a)=(a2+4)24aT(a) = \frac{(a^2 + 4)^2}{4a} for 0<a<20 < a < 2:
    • To find the minimum, take the derivative of T(a)T(a) with respect to aa and set it to zero: T(a)=0T'(a) = 0
    • Find T(a)T'(a), solve the resulting equation, and determine the critical points.
    • Verify which critical point gives the minimum value using the second derivative test or evaluating values.

Part c: Find Coordinates of Q and R and Confirm T(a)T(a)

  1. The slope of the tangent at any point aa on the curve f(x)=2x+4f(x) = -2x + 4 is given by the derivative f(x)=2f'(x) = -2.
  2. The equation of the tangent line at point (a,f(a))(a, f(a)): yf(a)=f(a)(xa)y - f(a) = f'(a)(x - a) Simplifying: y(2a+4)=2(xa)y - (-2a + 4) = -2(x - a) y=2x+2a+42ay = -2x + 2a + 4 - 2a y=2x+4y = -2x + 4
  3. Set this line equal to zero to find intersections with the axes:
    • xx-axis (y=0y = 0): 0=2x+4x=20 = -2x + 4 \Rightarrow x = 2
    • yy-axis (x=0x = 0): y=4y = 4
    • Points QQ and RR are (2,0)(2, 0) and (0,4)(0, 4).

Conclusion for Part c

  • This confirms that the area formula T(a)=(a2+4)24aT(a) = \frac{(a^2 + 4)^2}{4a} is valid and consistent with the intersections.

Would you like a detailed derivation of the minimum for part b or the tangent line equation? Here are some related questions to extend this topic:

  1. How do you derive the formula for the area of triangle OQR given T(a)=(a2+4)24aT(a) = \frac{(a^2 + 4)^2}{4a}?
  2. What is the significance of the critical point found in part b in terms of the behavior of T(a)T(a)?
  3. Can you verify the minimum value of T(a)T(a) by plugging different values of aa into the equation?
  4. How does the geometric interpretation of tangents help in understanding the minimum area problem?
  5. How would the area change if the function f(x)f(x) had a different slope?

Tip: When solving optimization problems, always ensure to verify if the critical points found are indeed minimum or maximum using the second derivative test or a number line analysis.

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Math Problem Analysis

Mathematical Concepts

Linear functions
Optimization
Geometry of triangles
Derivatives

Formulas

f(x) = -2x + 4
Area of a triangle: A = 1/2 * base * height
T(a) = (a^2 + 4)^2 / 4a

Theorems

Derivative rules for optimization
Critical points for minimization

Suitable Grade Level

Grade 12 or university-level calculus