Question 934802
{{{x}}}= length of one leg, in cm.
{{{y}}}= length of the other leg, in cm.
}}}According to the Pythagorean theorem,
{{{x^2+y^2=30^2}}}<--->{{{x^2+y^2=900}}}<--->{{{y^2=900-x^2}}}
The area of the right triangle, in {{{cm^2}}} is
{{{area=(1/2)*x*y}}},
so {{{area^2=((1/2)*x*y)^2}}}
{{{area^2=(1/2)^2*x^2*y*2}}}
{{{area^2=(1/4)*x^2*y*2}}}
Substituting the expression {{{900-x^2}}} (found above) for {{{y^2}}} we get
{{{area^2=(1/4)*x^2*(900-x^2)}}}<---->{{{area^2=(1/4)(-x^4+900x^2)}}}<---->{{{area^2=(1/4)(-(x^2)^2+900x^2)}}}
That last expression shows {{{area^2}}} a as quadratic function of {{{x^2)}}} ,
and we know that it will have a maximum when {{{x^2=900/2}}}<--->{{{x^2=450}}} .
For that value of {{{x^2}}} ,
{{{area^2=(1/4)(-450^2+900*450)}}}
{{{area^2=(1/4)*450*(-450+900)}}}
{{{area^2=(1/4)*450*450}}}
{{{area^2=(1/4)*450^2}}}
{{{area^2=450^2/2^2}}}
{{{area^2=(450/2)^2}}}
{{{area^2=225^2}}}
{{{highlight(area=225)}}} and that is the largest possible area of the triangle, in square centimeters.