The length of a rectangle is decreasing at the rate of 1 foot per second, but the area remains constant. How fast is the rectangle's width increasing when its length is 10 feet and its width is 5 feet?
A) $\frac { 1 } { 2 } \mathrm { ft } / \mathrm { s }$

B) $4 \mathrm { ft } / \mathrm { s }$
C) $\frac { 1 } { 10 } \mathrm { tt } / \mathrm { s }$
D) $2 \mathrm { ft } / \mathrm { s }$
E) $5 \mathrm { ft } / \mathrm { s }$
F) $\frac { 1 } { 4 } \mathrm { ft } / \mathrm { s }$
G) $10 \mathrm { ft } / \mathrm { s }$

H)$\frac { 1 } { 5 } \mathrm { tt } / \mathrm { s }$

Question

The length of a rectangle is decreasing at the rate of 1 foot per second, but the area remains constant. How fast is the rectangle's width increasing when its length is 10 feet and its width is 5 feet?
A) $\frac { 1 } { 2 } \mathrm { ft } / \mathrm { s }$

B) $4 \mathrm { ft } / \mathrm { s }$
C) $\frac { 1 } { 10 } \mathrm { tt } / \mathrm { s }$ 
D) $2 \mathrm { ft } / \mathrm { s }$ 
E) $5 \mathrm { ft } / \mathrm { s }$ 
F) $\frac { 1 } { 4 } \mathrm { ft } / \mathrm { s }$ 
G) $10 \mathrm { ft } / \mathrm { s }$

H)$\frac { 1 } { 5 } \mathrm { tt } / \mathrm { s }$

Quizplus · Accepted Answer

The area of the rectangle remains constant, so the rate of change of width must compensate for the rate of change of length to keep the area the same. Given the area $A = l \times w$, and the rate of change of length $\frac{dl}{dt} = -1 \, \text{ft/s}$, we find the rate of change of width $\frac{dw}{dt}$ when $l = 10 \, \text{ft}$ and $w = 5 \, \text{ft}$. Since $A$ is constant, $dA/dt = 0 = l \frac{dw}{dt} + w \frac{dl}{dt}$. Substituting the given values, $0 = 10 \frac{dw}{dt} - 5$, solving for $\frac{dw}{dt}$ gives $\frac{dw}{dt} = \frac{1}{2} \, \text{ft/s}$.

The Length of a Rectangle Is Decreasing at the Rate 12ft/s\frac { 1 } { 2 } \mathrm { ft } / \mathrm { s }21​ft/s

The Length of a Rectangle Is Decreasing at the Rate $\frac { 1 } { 2 } \mathrm { ft } / \mathrm { s }$