Asked by Megan
OLD MCDONALD IS GOING TO PUT A PIG PEN UP AGAINST HIS BARN. HE HAS 45 FEET OF FENCING AND WILL ONLY NEED TO MAKE 3 ADDITIONAL SIDES, AS THE BARN ACTS AS THE FOURTH SIDE.. WHAT IS THE MAXIMUM AREA THAT THE PIG PEN CAN HAVE.
Answers
Answered by
Scott
area = x * y
x + 2y = 45 ... x = 45 - 2y
substituting ... a = 45 y - 2 y^2
ymax = -b / 2a = -45 / (2 * -2) = 11.25
x = 22.5
a = 22.5 * 11.25 = 253.125
x + 2y = 45 ... x = 45 - 2y
substituting ... a = 45 y - 2 y^2
ymax = -b / 2a = -45 / (2 * -2) = 11.25
x = 22.5
a = 22.5 * 11.25 = 253.125
Answered by
Reiny
or,
by Calculus
short side --- x
single long side --- y
2x + y = 45
y = 45-2x
Area = A = xy = x(45-2x)
= 45x - 2x^2
dA/dx = 45 - 4x = 0 for a max of A
4x = 45
x = 45/4 , then y = 45-2(45/4) = 45/2
max area = xy = (45/4)(45/2)
= 2025/8
= appr 253.125 , same as Scott
by Calculus
short side --- x
single long side --- y
2x + y = 45
y = 45-2x
Area = A = xy = x(45-2x)
= 45x - 2x^2
dA/dx = 45 - 4x = 0 for a max of A
4x = 45
x = 45/4 , then y = 45-2(45/4) = 45/2
max area = xy = (45/4)(45/2)
= 2025/8
= appr 253.125 , same as Scott
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