Water reservoir

The cuboid reservoir contains 1900 hectoliters of water, and the water height is 2.7 m. Determine the bottom dimensions where one dimension is 2.2 m longer than the second.

Correct answer:

a =  7.36 m
b =  9.5605 m

Step-by-step explanation:

V=1900 hl→ m3=1900:10  m3=190 m3 c=2.7 m  V = abc S1=V/c=190/2.7=190027≐70.3704 m2 S1 = ab b = 2.2+a  a(2.2+a)=S1  a(2.2+a)=70.37037037037 a2+2.2a−70.37=0  p=1;q=2.2;r=−70.37 D=q2−4pr=2.22−4⋅1⋅(−70.37)=286.3214814815 D>0  a1,2=−q±D2p=−2.2±286.322 a1,2=−1.1±8.460518 a1=7.360518328 a2=−9.560518328  a=a1=7.3605=7.36 mV = 1900 \ \text{hl} \rightarrow \ \text{m}^3 = 1900 : 10 \ \ \text{m}^3 = 190 \ \text{m}^3 \ \\ c = 2.7 \ \text{m} \ \\ \ \\ V\ = \ abc \ \\ S_{1} = V/c = 190/2.7 = \dfrac{ 1900 }{ 27 } \doteq 70.3704 \ \text{m}^2 \ \\ S_{1}\ = \ ab \ \\ b\ = \ 2.2+a \ \\ \ \\ a(2.2+a) = S_{1} \ \\ \ \\ a(2.2+a) = 70.37037037037 \ \\ a^2 +2.2a -70.37 = 0 \ \\ \ \\ p = 1; q = 2.2; r = -70.37 \ \\ D = q^2 - 4pr = 2.2^2 - 4 \cdot 1 \cdot (-70.37) = 286.3214814815 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ -2.2 \pm \sqrt{ 286.32 } }{ 2 } \ \\ a_{1,2} = -1.1 \pm 8.460518 \ \\ a_{1} = 7.360518328 \ \\ a_{2} = -9.560518328 \ \\ \ \\ a = a_{1} = 7.3605 = 7.36 \ \text{m}

Our quadratic equation calculator calculates it.

b=a+2.2=7.3605+2.2≐9.5605 m   Verifying Solution:  V2=a⋅ b⋅ c=7.3605⋅ 9.5605⋅ 2.7=190 m3b = a + 2.2 = 7.3605 + 2.2 \doteq 9.5605 \ \text{m} \ \\ \ \\ \text{ Verifying Solution: } \ \\ V_{2} = a \cdot \ b \cdot \ c = 7.3605 \cdot \ 9.5605 \cdot \ 2.7 = 190 \ \text{m}^3

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