Can someone help me with this calculus problem?
Let f(x)=12-x^2 for x is greater or equal to zero and f(x) is greater or equal to 0.
(a) The line tangent to the graph of f(x) at the point (k, f(k)) intercepts the x-axis at x=4. What is the value of k?
(b)An isosceles triangle whose base is the interval from (0,0) to (c,0) has its vertex on the graph f(x). For what value of c does the triangle have maximum area? justify your answer.
THANK YOU SOO MUCH!
- etchLv 410 年前最佳解答
Part a:From the first statement of this problem, f(x) is onlyvalid in the first quadrant, for x between 0 and the square root of 12. Take the first derivative of f(x). You will get -2x , which is the slopeof the line.The line passes through the following two points(k,12-k^2) and (4,0) with a slope of –2k Therefore (12-k^2 – 0) = -2k (k – 4) Solve for k. You will find k = 2 or 6However, k is between 0 and the square root of 12.Therefore, k cannot be 6.Only k = 2 is the answer to Part a.
The vertex is on f(x) implies that the vertex of the isosceles is (c/2 , 12-(c/2)^2 )The area of the triangle is 1 / 2 * c *(12-(c/2)^2), which is 6c-(c^3)/8For the triangle to have maximum area, the derivative of (6c-(c^3)/8) = 0That is, 6-3/8 (c^2) = 0 Solve for c and you will find c = 4
- H. M. CLv 710 年前
This is not calculus, but analytic geometry 解析幾何, and relatively easy and simple ones .. Important for you to really learn these fundamental questions and answers ..
You need to plot them out [在座標上畫出來], so that you can see the right triangles 直角三角形 that allows you to relate the Ks with known values like, 4 and the f(x)=12-x^2, thus f(k)=12-k^2
Hope that helps you to do your work to figure this out ..