Two sides of a rectangle lie along the x and y axes. The vertex opposite the origin is in the first quadrant and lies on 3x + 2y = 6. What is the maximum area of the rectangle?

Question

Answer: 1.5

Thanks for any help

Answer 1

The horizonatal length is x and the vertical length is y.

Rearrange 3 x + 2 y = 6

2 y = 6 - 3 x

2 y = - 3 x + 6

y = ( - 3 x + 6 ) / 2

y = - 3 x / 2 + 3

A = x ∙ y = x ∙ ( - 3 x / 2 + 3 ) = - 3 x² / 2 + 3 x

dA / dy = y´= - 3 ∙ 2 x / 2 + 3 = - 3 x + 3

The first derivative test:

when y´= 0 a function has a extreme value

- 3 x + 3 = 0

Divide both sides by 3

- x + 1 = 0

- x = - 1

Multiply both sides by - 1

x = 1

y" = ( y´ )´

y" = - 3

Second derivative test:

If y" > 0 a function has a minimum

If y" < 0 a function has a maximum

y" = - 3 < 0

for x = 1 a function has a maximum

A = - 3 x² / 2 + 3 x

Amax = A (1) = - 3 ∙ 1² / 2 + 3 ∙ 1 = - 3 ∙ 1 / 2 + 3 =

- 3 / 2 + 3 = - 3 / 2 + 6 / 2 = 3 / 2 = 1.5

Answer 2

Wow, that's smart!!!