求曲线f(x)=4-x²上正坐标点(a,b)使围出三角形面积最小

Hey there, let's break this down step by step. I think you probably meant the triangle formed by the tangent line to the curve at ((a,b)) with the x-axis and y-axis—since the curve itself creates a curved region with the axes, not a triangle. That makes sense for the calculus approach you mentioned, so let's go with that!

Step 1: Use the point-on-curve condition #

Since ((a,b)) lies on (f(x)=4-x^2), we know:

b = 4 - a²

And since (a,b>0), this means (0 < a < 2) (when (a=2), (b=0) which doesn't meet our positive requirement).

Step 2: Find the tangent line at ((a,b)) #

First, take the derivative of (f(x)) to get the slope of the tangent at any point:

f'(x) = -2x

At (x=a), the slope is (f'(a) = -2a). Using point-slope form for the tangent line:

y - b = -2a(x - a)

Substitute (b=4-a²) and simplify:

y = -2a x + a² + 4

Step 3: Find intercepts of the tangent line with the axes #

• Y-intercept: Set (x=0), so (y = a² + 4) → intercept at ((0, a²+4))
• X-intercept: Set (y=0), solve for (x):

  0 = -2a x + a² + 4 → x = (a² + 4)/(2a)
  

  Intercept at (\left( \frac{a²+4}{2a}, 0 \right))

Step 4: Define the area function #

Since both intercepts are positive (thanks to (0 < a < 2)), the area (S(a)) of the triangle is half the product of the intercept lengths:

S(a) = (1/2) * \frac{a²+4}{2a} * (a²+4) = \frac{(a²+4)^2}{4a}

To make differentiation easier, rewrite this as:

S(a) = \frac{a³}{4} + 2a + \frac{4}{a}

Step 5: Find critical points by differentiating #

Take the derivative of (S(a)):

S'(a) = \frac{3a²}{4} + 2 - \frac{4}{a²}

Set (S'(a)=0) to find minimum candidates:

\frac{3a²}{4} + 2 - \frac{4}{a²} = 0

Multiply through by (4a²) to eliminate denominators (since (a≠0)):

3a⁴ + 8a² - 16 = 0

Let (u=a²) (where (u>0)), turning this into a quadratic equation:

3u² + 8u -16 =0

Solve using the quadratic formula (u = \frac{-b±\sqrt{b²-4ac}}{2a}):

u = \frac{-8 ± \sqrt{64 + 192}}{6} = \frac{-8 ±16}{6}

We only care about the positive root:

u = \frac{8}{6} = \frac{4}{3} → a²=\frac{4}{3} → a=\frac{2\sqrt{3}}{3}

Step 6: Verify it's a minimum and find (b) #

The second derivative of (S(a)) is:

S''(a) = \frac{6a}{4} + \frac{8}{a³}

For (a>0), (S''(a) >0), so this critical point is a local minimum—and since it's the only critical point in (0 < a <2), it's the global minimum.

Now find (b):

b =4 - a² =4 - \frac{4}{3} = \frac{8}{3}

Final Result #

The point ((a,b)) that minimizes the tangent triangle area is (\left( \frac{2\sqrt{3}}{3}, \frac{8}{3} \right)), with the minimum area being (\frac{32\sqrt{3}}{9}).

内容的提问来源于stack exchange，提问作者dirkpra