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# Rate of Change With Derivatives Problem?

I think I found a solution for this problem, but I don't think its right.

The area of a circular segment (crust of a piece of pizza) is given by the formula (1/2) *r^2 * (theta - sin(theta)) where r is the radius and theta is the central angle of the circle in radians. Also, s= r*theta, where s is the circular arc. If the perimeter of the circular sector (entire piece of pizza) is fixed at 200 ft, what valuse of r and s will give the segment the greatest area.

BTW, this is the exact question as it is on my worksheet. I got confused by perimeter of circular sector (does that include the sides or just the arc).

### 2 Answers

- ?Lv 51 decade agoFavorite Answer
The perimeter includes the sides

The perimeter is p = 2r + s = 2r + r θ = 200

then θ = (200 – 2r)/r

The area of the sector is

A(r) = ½ r s = ½ r² θ = [r(200 – 2r)]/2 = - r² + 100r

A’(r) = - 2r + 100 = 0 for r = 50

The area is the greatest when r = 50 ft

and s = 200 – 2*50 = 100 ft

- Anonymous4 years ago
placed the sunshine at a distance D = 30 ft to the wall, enable Mona be T = 6 ft tall, enable her shadow be s ft tall. denote Mona's distance to the sunshine be x(t) the place t = time. Her velocity v is then: dx/dt = v Now in case you comedian strip this out you have 2 desirable triangles first one has x as a base and T as a leg 2nd one has L as a base and s as a leg the backside of the two triangles lay on mind-blowing of one yet another, and since the shadow, s, is defined by the line made by the sunshine, the mind-blowing of Mona's head, and the element on the wall the sunshine reaches, the two triangle's hypotoneus lay on choose of one yet another, so as that they've a similar indoors angles. call the attitude between the hypotoneus and base q. Then by elementary trig: tan(q) = T/x = s/D ---> x = TD/s Take d/dt for locate velocity: v = dx/dt = -TD/s^2 ds/dt Now you're instructed ds/dt = 3 ft/s whilst s = 9 ft so v = -6ft*30ft*3ft/s/(9 ft)^2 = - 2ft/sec the minus sign shows she is working in the direction of the sunshine because we defined the gap from the sunshine to th ewall as a favorable displacement (path).