Find Min+Max

Geometry Level 2

y = sin ⁡ 2 x sin ⁡ x + 8 y = \frac {\sin^2 x}{\sin x + 8}

For y y as defined above, find min ⁡ ( y ) + max ⁡ ( y ) \min(y) + \max(y) .


The answer is 0.1428.

Aly Ahmed by Aly Ahmed , 34, Egypt

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2 solutions

Richard Desper
Jan 14, 2020

Easier question: replace sin ⁡ ( x ) \sin(x) with x x , and let it range over [ − 1 , 1 ] [-1,1] . Then this is a first-year Calculus problem. Nothing about the behavior of the sine function is reflected in the calculation of the extremal values of the given function other than the fact that it (the sine function) takes on all values in the range from -1 to 1.

1 Helpful 1 Interesting 1 Brilliant 0 Confused
Chew-Seong Cheong
Jan 14, 2020

Since 0 ≤ sin ⁡ 2 x ≤ 1 0 \le \sin^2 x \le 1 and 7 ≤ sin ⁡ x + 8 ≤ 9 7 \le \sin x + 8 \le 9 , y ≥ 0 y \ge 0 and min ⁡ ( y ) = 0 \min(y) = 0 , when sin ⁡ x = 0 \sin x = 0 . We note that max ⁡ ( y ) = max ⁡ ( sin ⁡ 2 x ) min ⁡ ( sin ⁡ x + 8 ) = 1 7 \max(y) = \dfrac {\max(\sin^2 x)}{\min(\sin x + 8)} = \dfrac 17 , when sin ⁡ x = − 1 \sin x = -1 . Therefore, min ⁡ ( y ) + max ⁡ ( y ) = 0 + 1 7 ≈ 0.143 \min (y) + \max (y) = 0 + \dfrac 17 \approx \boxed{0.143} .

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Why can't we use calculus in it. On finding critical points and then finding the image in f(x) may give the answer. I tried to solve in this way and I ended up at maximum on π/2 and minimum on 3π/2. My maximum was 1\9 and minimum as -1/7. Where I got wrong. Please let me know soon.

Pradeep Tripathi - 1 year, 4 months ago

@Pradeep Tripathi , as mentioned in my solution sin ⁡ 2 x ≥ 0 \sin^2 x \ge 0 and sin ⁡ x + 8 ≥ 0 \sin x + 8 \ge 0 , f ( x ) f(x) is always positive f ( x ) = − 1 7 f(x) = - \frac 17 is wrong.

f ( x ) = sin ⁡ 2 x sin ⁡ x + 8 f ′ ( x ) = 2 sin ⁡ x cos ⁡ x ( sin ⁡ x + 8 ) − sin ⁡ 2 x cos ⁡ x ( sin ⁡ x + 8 ) = sin ⁡ x cos ⁡ x ( sin ⁡ x + 16 ) ( sin ⁡ x + 8 ) 2 \begin{aligned} f(x) & = \frac {\sin^2 x}{\sin x + 8} \\ f'(x) & = \frac {2\sin x \cos x(\sin x+8)-\sin^2 x \cos x}{(\sin x + 8)} \\ & = \frac {\sin x \cos x(\sin x + 16)}{(\sin x + 8)^2} \end{aligned}

f ′ ( x ) = 0 f'(x)=0 , when sin ⁡ x = 0 \sin x = 0 , then f ( x ) = 0 f(x) = 0 , the minimum; and also when cos ⁡ x = 0 ⟹ x = ± π 2 \cos x = 0\implies x = \pm \frac \pi 2 . For x = π 2 x = \frac \pi 2 , f ( x ) = 1 9 f(x) = \frac 19 and for x = − π 2 x = -\frac \pi 2 , f ( x ) = 1 7 f(x) = \frac 17 , the maximum.

Chew-Seong Cheong - 1 year, 4 months ago

Oh! You are correct. It was a calculation mistake that I felt nostalgic about it. Thanks to spent some of your crucial time to answer me.

Pradeep Tripathi - 1 year, 4 months ago

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