A wire of length 25 m is to be cut into two pieces. One of the p

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 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. 

120 Views

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.

75 Views

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

150 Views

 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.
75 Views

Advertisement
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.

114 Views

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.
438 Views

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.

80 Views

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.
101 Views

Advertisement
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?

86 Views

 Multiple Choice QuestionsShort Answer Type

Advertisement

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?


Total length of wire = 25 metre. Let x metres be made into a circle and (25 – x) metres into a square.
space therefore space space space space space radius space of space circle space equals space fraction numerator straight x over denominator 2 straight pi end fraction metres space and space side space of space square space space equals space fraction numerator 25 minus straight x over denominator 4 end fraction metres
space Let space straight S space be space the space sum space of space areas space of space two space figures. space
therefore space space space space space space space straight S space equals space straight pi open parentheses fraction numerator straight x over denominator 2 straight pi end fraction close parentheses squared plus open parentheses fraction numerator 25 minus straight x over denominator 4 end fraction close parentheses squared space equals space fraction numerator straight x squared over denominator 4 straight pi end fraction plus fraction numerator left parenthesis 25 minus straight x right parenthesis squared over denominator 16 end fraction
space space space space space space space space space dS over dx equals fraction numerator 2 straight x over denominator 4 straight pi end fraction plus fraction numerator 2 left parenthesis 25 minus straight x right parenthesis thin space left parenthesis negative 1 right parenthesis over denominator 16 end fraction space equals space fraction numerator straight x over denominator 2 straight pi end fraction minus fraction numerator 25 minus straight x over denominator 8 end fraction
space space space space space space space space space space space space dS over dx equals 0 space space gives space us space space fraction numerator straight x over denominator 2 straight pi end fraction minus fraction numerator 25 minus straight x over denominator 8 end fraction equals 0 space

therefore space space space space 4 straight x minus 25 straight pi space plus space xπ space equals space 0 space space space space space space rightwards double arrow space space space space left parenthesis straight pi plus 4 right parenthesis space straight x space equals space 25 straight pi

therefore space space space space space space straight x space equals space fraction numerator 25 space straight pi over denominator straight pi plus 4 end fraction

          fraction numerator straight d squared straight S over denominator dx squared end fraction space equals fraction numerator 1 over denominator 2 straight pi end fraction plus 1 over 8

When straight x space equals fraction numerator 25 space straight pi over denominator straight pi space plus 4 end fraction comma space space space space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space fraction numerator 1 over denominator 2 straight pi end fraction plus 1 over 8 greater than 0
therefore space space space space space straight S space is space minimum space when space straight x space equals fraction numerator 25 space straight pi over denominator straight pi plus 4 end fraction
therefore space space space the space wire space is space to space be space cut space at space straight a space distance space of space fraction numerator 25 space straight pi over denominator straight pi plus 4 end fraction metres space from space one space end. space
75 Views

Advertisement
Advertisement