A window is in the form of a rectangle surmounted by a semi-circ

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 Multiple Choice QuestionsShort Answer Type

301.

Manufacturer can sell x items at a price of rupees open parentheses 5 minus straight x over 100 close parentheses space each. The cost price of x items is Rs. open parentheses straight x over 5 plus 500 close parentheses. Find the number of items he should sell to earn maximum profit. 

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302.

A firm has found from experience that its profit as a function of x, the amount produced, is given by
straight p left parenthesis straight x right parenthesis space equals space minus straight x cubed over 3 plus 729 space straight x space minus 2500 comma space space space 0 space less or equal than space straight x space less or equal than space 35.
Find the value of x that maximises the profit.

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303.

A beam of length l is supported at one end. If W is the uniform load per unit length, the bending moment M at a distance x from the end is given by straight M space equals 1 half space straight l space straight x space space minus space 1 half space straight W space straight x squared. maximum value. Find the point on the beam at which the bending moment has th

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 Multiple Choice QuestionsLong Answer Type

304. Given the sum of the perimeters of a square and a circle, prove that the sum of their areas is least when the side of the square is equal to the diameter of the circle.
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305.

A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.

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306. A figure consists of a semi-circle with a rectangle on its diameter. Given perimeter of the figure, find the dimensions in order that the area may be maximum.
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307.

A window consists of a semi-circle with a rectangle on its diameter. If the perimeter of the window is 30 metres, find the dimensions of the window in order that its area may be maximum.
Or
A window is in the form of a rectangle surmounted by a semi-circle. If the total perimeter of the window is 30 m. find the dimensions of the window so that maximum light is admitted.

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308. A window is in the form of a rectangle surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.


Let x, y be length, breadth of the rectangle ABCD.
Let straight x over 2 be the radius of the semi-circle with centre at O.
Perimeter of figure  = 10 meters
therefore space space space space straight x plus 2 straight y plus straight pi straight x over 2 space equals space 10 space space space space space space space rightwards double arrow space space 2 straight x plus 4 straight y plus πx space equals 20
therefore space space space space space space 4 straight y space equals 20 minus left parenthesis straight pi plus 2 right parenthesis straight x
rightwards double arrow space space space space space straight y space equals space fraction numerator 20 minus left parenthesis straight pi plus 2 right parenthesis space straight x over denominator 4 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space space space space space
Let A be the area of the figure.
       therefore space space space straight A space equals space xy space plus 1 half straight pi open parentheses straight x over 2 close parentheses squared space equals straight x open square brackets fraction numerator 20 minus left parenthesis straight pi plus 2 right parenthesis space straight x over denominator 4 end fraction close square brackets plus πx squared over 8               open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

     therefore space space space space straight A space equals space left square bracket 20 straight x minus left parenthesis straight pi plus 2 right parenthesis straight x squared right square bracket space plus πx squared over 8
therefore space space space dA over dx space equals 1 fourth left square bracket 20 minus space 2 left parenthesis straight pi plus 2 right parenthesis straight x right square bracket plus space πx over 4
Now space space dA over dx equals 0 space space space space space rightwards double arrow space space space space 1 fourth left square bracket 20 minus 2 left parenthesis straight pi minus 2 right parenthesis straight x right square bracket plus πx over 4 equals 0
rightwards double arrow space space space space space 20 minus space 2 left parenthesis straight pi plus 2 right parenthesis space straight x space plus space πx space equals space 0
rightwards double arrow space space space space space space 20 minus 2 πx minus 4 straight x plus πx space equals space 0
rightwards double arrow space space space space space 20 minus left parenthesis straight pi plus 4 right parenthesis straight x space equals space 0 space space space space rightwards double arrow space space space straight x space equals space fraction numerator 20 over denominator straight pi plus 4 end fraction
                   fraction numerator straight d squared straight A over denominator dx squared end fraction space equals space 1 fourth left square bracket 0 minus 2 left parenthesis straight pi plus 2 right parenthesis right square bracket space plus straight pi over 4 space equals 1 fourth left square bracket negative 2 straight pi minus 4 plus straight pi right square bracket space equals negative fraction numerator straight pi plus 4 over denominator 4 end fraction

When space straight x space equals fraction numerator 20 over denominator straight pi plus 4 end fraction. space fraction numerator straight d squared straight A over denominator dx squared end fraction space equals negative fraction numerator straight pi plus 4 over denominator 4 end fraction less than 0
therefore space space space space space space space space space space straight A space is space maximum space when space straight x space equals fraction numerator 20 over denominator straight pi plus 4 end fraction
and   straight y space equals space fraction numerator 20 minus left parenthesis straight pi plus 2 right parenthesis space begin display style fraction numerator 20 over denominator straight pi plus 4 end fraction end style over denominator 4 end fraction space equals space fraction numerator 20 straight pi plus 80 minus 20 straight pi minus 40 over denominator 4 left parenthesis straight pi plus 4 right parenthesis end fraction space equals space fraction numerator 10 over denominator straight pi plus 4 end fraction
therefore   area of figure is maximum i.e., maximum light is admitted when length of rectangle = straight x equals fraction numerator 20 over denominator straight pi plus 4 end fraction comma space space breadth space of space rectangle space equals straight y space equals space fraction numerator 10 over denominator straight pi plus 4 end fraction comma
radius space of space semi minus circle space equals space straight x over 2 space equals space fraction numerator 5 over denominator straight pi plus 4 end fraction.  
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309.

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

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 Multiple Choice QuestionsShort Answer Type

310. A wire of length 25 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What would be the lengths of the two pieces, so,that combined area of the square and the circle is minimum?
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