The area of a trapezium is defined by function f and given by$$f(x)=(10+x)\sqrt{100}−x^2$$, then the area when it is maximised is:

(a) 75cm2

(b) $$7\sqrt{3}cm^2$$

(c) $$75\sqrt{3}cm^2$$

(d) 5cm2

Asked by Abhisek | 1 year ago |  124

#### 1 Answer

##### Solution :-

Right answer is (c) $$75\sqrt{3}cm^2$$

Explanation:-

$$f′(x)=\dfrac{-2x^2-10x+100}{\sqrt{100-x^2}}$$

$$f′(x)=0 =x=−10$$

$$or\; 5 , But \;x>0=x=5$$

$$f''(x)=\dfrac{2x^3-300x-1000}{(100-x)^\dfrac{3}{2}}$$

$$f''(5)=\dfrac{-30}{\sqrt{75}}<0$$

Maximum area of trapezium is $$75\sqrt{3}cm^2$$ when x = 5

Answered by Pragya Singh | 1 year ago

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