Area of Largest Rectangle in 1st Quadrant of y=4-x^2

[fitz_calc]

Homework Statement

area of largest rectangle with sides parallel to axes (1st quad) under y=4-x^2

The Attempt at a Solution

x(4-x^2)

Area ` = (4-3x^2)

x=0 and sqrt(4/3)

The book says 4/3 (1.15) is wrong and that 3.08 is correct, how is this possible?

 

[rs1n]

Always draw a diagram whenever possible. Your area formula isn't quite correct. Moreover, your derivative isn't correct (either use the product rule, or first expand your formula for the area and then differentiate). But most importantly, once you find critical points always check to see if your critical points are indeed max/mins via either the first or second derivative test!

 

[HallsofIvy]

Homework Statement

area of largest rectangle with sides parallel to axes (1st quad) under y=4-x^2

The Attempt at a Solution

x(4-x^2)

Area ` = (4-3x^2)

x=0 and sqrt(4/3)

The book says 4/3 (1.15) is wrong and that 3.08 is correct, how is this possible?

Why would you think (4/3)(1.15) would be correct? For $x= \sqrt{4/3}$, that is (x²)(x), NOT x(4- x²).

rs1n, perhaps the original post has been edited since you saw it but that IS the correct area (since the rectangle is only in the first quadrant) and the derivative is correct. He got the correct value for x but then calculated the area incorrectly.

 

[rs1n]

Why would you think (4/3)(1.15) would be correct? For $x= \sqrt{4/3}$, that is (x²)(x), NOT x(4- x²).

rs1n, perhaps the original post has been edited since you saw it but that IS the correct area (since the rectangle is only in the first quadrant) and the derivative is correct. He got the correct value for x but then calculated the area incorrectly.

No, I was quite mistaken (because I've used this problem on an exam before, but with the upper half plane, and just assumed it was the same :-). Thank you for pointing out my mistake. You are also correct in pointing out that his derivative is correct. Wow, what a lapse in thinking for me on this one!