Let x, y be non-negative numbers with
sqrt(1-x^2/4)+sqrt(1-y^2/16)=3/2.
Find the maximum value of xy.
First solve for y in terms of x:
>> syms x y
>> eq=sqrt(1-x^2/4)+sqrt(1-y^2/16)==3/2
eq =
(1 - x^2/4)^(1/2) + (1 - y^2/16)^(1/2) == 3/2
>> solve(eq,y)
Warning: Solutions are only valid under certain conditions. To include parameters
and conditions in the solution, specify the 'ReturnConditions' value as 'true'.
> In sym/solve>warnIfParams (line 478)
In sym/solve (line 357)
ans =
-8*(x^2/16 + (3*(1 - x^2/4)^(1/2))/4 - 9/16)^(1/2)
8*(x^2/16 + (3*(1 - x^2/4)^(1/2))/4 - 9/16)^(1/2)
Taking the positive case, tabulate and narrow down:
clearvars, clc
xy=0;
for x=0:.001:4
prev=xy;
y= 8*(x^2/16 + (3*(1 - x^2/4)^(1/2))/4 - 9/16)^(1/2);
xy=x*y;
if isreal(xy)
disp([x xy xy-prev])
end
end
ultimately gets narrowed down (via where xy - prev changes sign) to
clearvars, clc
xy=0;
for x= 1.32287565
prev=xy;
y= (8)*(x^2/16 + (3*(1 - x^2/4)^(1/2))/4 - 9/16)^(1/2);
xy=x*y;
if isreal(xy)
disp([(x) y (xy) (xy-prev)])
end
end
1.32287565 2.64575132212918 3.5 3.5
Wolfram Alpha suggests sqrt(7)/2 for 1.32287565.
If that's true so that x = sqrt(7)/2, then y = sqrt(7) ~= 2.645751311064591. In any case, maximum xy = 3.5 = 7/2.
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Posted by Charlie
on 2021-12-14 11:07:53 |
computer guided solution
