Let f(x) = 12 - x^2 for x≥0 and f(x) ≥ 0
(a) The line tangent to the graph of f 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 of f. For what value of c does the triangle have maximum area? Justify your answer.

Please show work because I need to check my work. Thank you.

Question

Let f(x) = 12 - x^2 for x≥0 and f(x) ≥ 0
(a) The line tangent to the graph of f 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 of f. For what value of c does the triangle have maximum area? Justify your answer.

Please show work because I need to check my work. Thank you.

nikvist · Answer

$$f(x)=12-x^2\quad;\quad x\geq 0\quad,\quad f(x)\geq 0$$a.$$f'(x)=-2x$$$$y-f(k)=-2k(x-k)$$$$-f(k)=-2k(4-k)$$$$k^2-12=-8k+2k^2$$$$k^2-8k+12=(k-2)(k-6)=0\quad\Rightarrow\quad k=2$$b.$$S=\frac{1}{2}cf(c/2)=\frac{c}{2}\left(12-\frac{c^2}{4}\right)=6c-\frac{c^3}{8}$$$$S'=6-\frac{3c^2}{8}=3\left(2-\frac{c^2}{8}\right)=0\quad\Rightarrow\quad c=4$$$$S_{\max}=6\cdot 4-\frac{4^3}{8}=24-8=16$$