An animal that dives into a pool of water is at a depth of x, in feet,

There are two ways of doing this problem. The first one is for people who still remembers what derivatives are and the second one is applying logic to know after how many seconds the animal reaches the geatest depth

1º DERIVATIVES

Engineers taking the GMAT will still remember that to calculate the max/min value of a function f(x) you must equal to zero the derivative of f(x) (denoted by f'(x))

derivate of f(x) = −4(t−2)^2+16 ---> f'(x) = -4*2(t-2) = 0 ---> t = 2 (this means that after 2 seconds the animal reached the highest depth)

so we need to substitute t = 2 + 0.5 in the original equation to know the depth of the animal 0.5 seconds after reaching the greatest depth

x = −4(t−2)^2+16 ---> x = -4(2.5-2)^2+16 ---> x = -4(0.5)^2+16 ---> x = -1+16 ---> x = 15

OPTION C

2º CALULATING t WHEN REACHING GREATEST DEPTH

The equation is x = −4(t−2)^2+16 and we need to make x the highest possible. When would it be the highest taking into account that one term is adding up (+16) and one term deducting (−4(t−2)^2)?. The answer is when the deducting element is zero so that x is max. The deducting term is zero when t = 2

Therefore, we need to calculate x when t is 2 + 0.5 = 2.5 seconds

Again solution is x = 15

OPTION C

Given: An animal that dives into a pool of water is at a depth of x, in feet, after t seconds, where \(x = - 4(t - 2)^2 + 16\).

Asked: At what depth, in feet, is the animal 0.5 seconds after it reaches its greatest depth?

An animal that dives into a pool of water is at a depth of x, in feet, after t seconds, where \(x = - 4(t - 2)^2 + 16\).
x is max when t=2 deepest depth x = 16

At what depth, in feet, is the animal 0.5 seconds after it reaches its greatest depth
At t = 2+.5 =2.5
\(x= -4(.5)^2 +16 = 16 - 4*.25 = 16 -1 = 15\)

IMO C

Read the equation again it is -4(t -2)^2 +16 …… the highest depth can be 16 when t = 2.

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