Total length of the wire = lNow, l = length of wire bent in the form of circle + length of wire bent in the form of square⇒l = 2πr + 4x⇒x = l - 2πr4 1Let A denotes the sum of the areas of circle and square A = πr2 + x2⇒A = πr2 + l - 2πr42⇒A = πr2 + l2 + 4π2r2 - 4lπr16⇒dAdr = 2πr + 8π2r - 4lπ16⇒dAdr = 2πr + 2π2r - lπ4For maxima or minima, dAdr = 0⇒2πr + 2π2r - lπ4 = 0⇒2πr = lπ - 2π2r4⇒8πr = lπ - 2π2r⇒8r = l - 2πr⇒8r = 2πr + 4x - 2πr⇒8r = 4x⇒r = x2Now, d2Adr2 = 2π + 2π24 =2π + π22Now, d2Adr2r=x2 = 2π + π22 > 0So, A is least when r = x2 or x = 2r

Pls answer q23

Pls answer q23 24. Its 25. one part square is least. if 2 The value of a for which the sum of e squares the roots of the equation — (a —2) x — a + — least, is (B) Two towns A and B are 60 km apart. A school be built to serve 150 students in town A students in town B. If the total distance travelled by all 200 students is to be as possible, then the school should be built at (A) town B

Total length of the wire = l Now, l = length of wire bent in the form of circle + length of wire bent in the form of square \Rightarrow l = 2 πr + 4 x \Rightarrow x = \frac{l - 2 πr}{4} . . . . (1) Let A denotes the sum of the areas of circle and square . A = {πr}^{2} + x^{2} \Rightarrow A = {πr}^{2} + {[\frac{l - 2 πr}{4}]}^{2} \Rightarrow A = {πr}^{2} + [\frac{l^{2} + 4 π^{2} r^{2} - 4 lπr}{16}] \Rightarrow \frac{dA}{dr} = 2 πr + [\frac{8 π^{2} r - 4 lπ}{16}] \Rightarrow \frac{dA}{dr} = 2 πr + (\frac{2 π^{2} r - lπ}{4}) For maxima or minima, \frac{dA}{dr} = 0 \Rightarrow 2 πr + (\frac{2 π^{2} r - lπ}{4}) = 0 \Rightarrow 2 πr = \frac{lπ - 2 π^{2} r}{4} \Rightarrow 8 πr = lπ - 2 π^{2} r \Rightarrow 8 r = l - 2 πr \Rightarrow 8 r = 2 πr + 4 x - 2 πr \Rightarrow 8 r = 4 x \Rightarrow r = \frac{x}{2} Now, \frac{d^{2} A}{{dr}^{2}} = 2 π + \frac{2 π^{2}}{4} = 2 π + \frac{π^{2}}{2} Now, {[\frac{d^{2} A}{{dr}^{2}}]}_{r = \frac{x}{2}} = 2 π + \frac{π^{2}}{2} > 0 So, A is least when r = \frac{x}{2} or x = 2 r

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