SOLUTION: Determine the equation of the line tangent to 9x^2 + 16y^2 = 144 forming the triangle of least area with the x and y axes in the first qudrant. What is the area of the triangle wi

Algebra ->  Surface-area -> SOLUTION: Determine the equation of the line tangent to 9x^2 + 16y^2 = 144 forming the triangle of least area with the x and y axes in the first qudrant. What is the area of the triangle wi      Log On


   

Question 1183554: Determine the equation of the line tangent to 9x^2 + 16y^2 = 144 forming the triangle of least area with the x and y axes in the first qudrant. What is the area of the triangle with least area?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2F16+%2B+y%5E2%2F9+=+1 ===> x%2F16+%2B+%28y%2A%28dy%2Fdx%29%29%2F9=0 ===> dy%2Fdx+=+-%289x%29%2F%2816y%29.
Now let (x%5B0%5D, y%5B0%5D) be the point of tangency in the 1st quadrant. The tangent line should be
y+-+y%5B0%5D+=+-%28%289x%5B0%5D%29%2F%2816y%5B0%5D%29%29%2A%28x-x%5B0%5D%29.
The y-intercept of this is y+=+9%2Fy%5B0%5D. (Verify!)
And the x-intercept is x+=+16%2Fx%5B0%5D. (Verify!)
The area of the triangle in the 1st quadrant is then A+=+72%2F%28x%5B0%5D%2Ay%5B0%5D%29+=+96%2F%28x%5B0%5D%2Asqrt%2816+-+x%5B0%5D%5E2%29%29.
Now get the derivative of A wrt x%5B0%5D, and set it to 0:

===> x%5B0%5D+=+2sqrt%282%29, so there is a local extremum at this point.
If x+%3C+2sqrt%282%29, then dA%2Fdx%5B0%5D+%3C+0. (E.g., choose x=sqrt%282%29.)
If x+%3E+2sqrt%282%29, then dA%2Fdx%5B0%5D+%3E+0. (E.g., choose x=3.)
Hence there is absolute minimum for the area in the 1st quadrant at x%5B0%5D+=+2sqrt%282%29, by the 1st derivative test.
The corresponding y-value is y%5B0%5D+=+%283sqrt%282%29%29%2F2.
The equation of the tangent line is then y+-+y%5B0%5D+=+-%28%289x%5B0%5D%29%2F%2816y%5B0%5D%29%29%2A%28x-x%5B0%5D%29, or highlight%28y+-+%283sqrt%282%29%29%2F2+=+-%283%2F4%29%2A%28x-2sqrt%282%29%29%29.

The area of the triangle of least area is A+=+72%2F%28x%5B0%5D%2Ay%5B0%5D%29+=+highlight%2812%29