A Course of Elementary Mathematics Affording Aid to Candidates for Admission I

J R John Radford Young

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And a min. , it is necessary to notice whether or not the sign oi f^(x) has been changed when equating it to zero : the expression represented by u above having been transposed, the sign of it has been changed. [In next prob. The sign will be pre- served. ] 284 SCHOLIUM.
Peob. 9. To determine when the area of the part of the leaf turned down, as in last problem, is the least possible.
It is plain that B^B is bisected at right angles by PQ, therefore PQ . B'B=A: times the area of B'PQ. Now by la...st prob. , PQz=. /- and B'B=.^2ax, . '. Multiplying the squares of ttese together, we have V 2iXa '7* 2 y* ~ CL 4a- =min. , . -. ;^=max. =M, . '. /{x)=2xaux*=0, . : f^{x)=2 iux^=0.
2xa X 3 2a Multiply this by -, and subtract, . *. -cc a=0, . -. X=. , and since /2(a7)= 12m^^ is negative, m, for this value of x^ is a maximum ; and therefore the proposed expression is a minimum.
Examples for Exercise.
(1) Of all right-angled plane triangles baring the same hypotenuse, to find that which has the greatest area.

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